Aimathic – Infinite limits and asymptotes

Infinite limits and asymptotes

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Difficulty: 1 – 5

54261812

Let $f(x)=\arctan x$. Find the limits of $f(x)$ as $x\to\infty$ and as $x\to-\infty$. Then state all horizontal asymptotes of the graph.

Hints

- Think about the output values that the inverse tangent function can approach but never exceed. - Consider the two ends of the x-axis separately; they need not approach the same height.

Solution

1. The range of $\arctan x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$, and $\arctan x$ approaches its upper endpoint as $x\to\infty$. 2. Therefore $\lim_{x\to\infty}\arctan x=\frac{\pi}{2}$, giving the horizontal asymptote $y=\frac{\pi}{2}$ on the right. 3. As $x\to-\infty$, $\arctan x\to-\frac{\pi}{2}$, giving the horizontal asymptote $y=-\frac{\pi}{2}$ on the left.

Answer

$\lim_{x\to\infty}\arctan x=\frac{\pi}{2}$ and $\lim_{x\to-\infty}\arctan x=-\frac{\pi}{2}$. The horizontal asymptotes are $y=\frac{\pi}{2}$ and $y=-\frac{\pi}{2}$.

54262812

A function $h$ satisfies $\lim_{x\to-2^-}h(x)=+\infty$, $\lim_{x\to-2^+}h(x)=+\infty$, $\lim_{x\to\infty}h(x)=4$, and $\lim_{x\to-\infty}h(x)=-1$. State every vertical and horizontal asymptote that is guaranteed by this information.

Hints

- Match unbounded behavior near a finite input with one type of asymptote. - Match finite limiting heights at the ends of the x-axis with another type of asymptote.

Solution

1. Both one-sided limits are unbounded as $x\to-2$, so $x=-2$ is a vertical asymptote. 2. Since $h(x)\to4$ as $x\to\infty$, $y=4$ is a horizontal asymptote on the right. 3. Since $h(x)\to-1$ as $x\to-\infty$, $y=-1$ is a horizontal asymptote on the left.

Answer

Vertical asymptote: $x=-2$. Horizontal asymptotes: $y=4$ as $x\to\infty$ and $y=-1$ as $x\to-\infty$.

52290412

Evaluate each one-sided limit. a) $\lim_{x\to1^-}\frac{x+2}{x-1}$ b) $\lim_{x\to-2^+}\frac{3}{x+2}$

Hints

- Determine the sign of the numerator and denominator near the approach value. - Track whether the denominator approaches zero from the positive or negative side.

Solution

1. In part a, the numerator approaches $3$, while the denominator approaches $0$ through negative values. Therefore, the quotient approaches $-\infty$. 2. In part b, the numerator is positive and the denominator approaches $0$ through positive values. Therefore, the quotient approaches $+\infty$.

Answer

a) $-\infty$ b) $+\infty$

52614012

A hypothetical savings account earns $100\%$ annual interest. If the year is divided into $n$ equal compounding periods, an initial balance of $\$100.00$ grows after one year to $K_n=100\left(1+\frac{1}{n}\right)^n$ dollars. 1. Find the ending balance for annual compounding $(n=1)$, monthly compounding $(n=12)$, and daily compounding $(n=365)$. Round to the nearest cent. 2. Find $\lim_{n\to\infty}K_n$, representing continuous compounding. 3. Find the difference between the daily-compounding balance and the continuous-compounding balance.

Hints

- Substitute each value of $n$ into the model. - Recall the limit that defines $e$. - Keep extra decimal places until the final subtraction.

Solution

1. $K_1=100(2)=\$200.00$. 2. $K_{12}=100\left(1+\frac{1}{12}\right)^{12}\approx\$261.30$. 3. $K_{365}=100\left(1+\frac{1}{365}\right)^{365}\approx\$271.46$. 4. Since $\left(1+\frac{1}{n}\right)^n\to e$, $K_n\to100e\approx\$271.83$. 5. The difference is $100e-K_{365}\approx\$0.37$.

Answer

1. $\$200.00$, $\$261.30$, and $\$271.46$ 2. $100e\approx\$271.83$ 3. Approximately $\$0.37$

52635512

Find the limits as $x\to\infty$ and as $x\to-\infty$. a) $f(x)=5-\frac{4}{x^2}$ b) $g(x)=\frac{3(4^x)+2}{4^x}$ c) $h(x)=8(0.5)^x-2$

Hints

- Rewrite quotients when possible. - Recall the behavior of exponential functions with bases greater than $1$ and between $0$ and $1$.

Solution

1. In part a, $\frac{4}{x^2}\to0$ in both directions, so $f(x)\to5$. 2. Rewrite part b as $g(x)=3+\frac{2}{4^x}$. As $x\to\infty$, the fraction approaches $0$, so $g(x)\to3$. As $x\to-\infty$, $4^x\to0^+$, so the fraction approaches $+\infty$, and $g(x)\to+\infty$. 3. In part c, $(0.5)^x\to0$ as $x\to\infty$, so $h(x)\to-2$. As $x\to-\infty$, $(0.5)^x\to+\infty$, so $h(x)\to+\infty$.

Answer

a) $\lim_{x\to\pm\infty}f(x)=5$ b) $\lim_{x\to\infty}g(x)=3$; $\lim_{x\to-\infty}g(x)=+\infty$ c) $\lim_{x\to\infty}h(x)=-2$; $\lim_{x\to-\infty}h(x)=+\infty$

52635712

Let $f(x)=\frac{8x+12}{4x-5}$. Find $\lim_{x\to\infty}f(x)$, and briefly explain its geometric meaning for the graph.

Hints

- Divide by the highest power of $x$. - Connect a finite limit at infinity with a horizontal asymptote.

Solution

1. Divide the numerator and denominator by $x$: $f(x)=\frac{8+\frac{12}{x}}{4-\frac{5}{x}}$. 2. As $x\to\infty$, the reciprocal terms approach $0$, so the limit is $\frac{8}{4}=2$. 3. Geometrically, the graph approaches the horizontal line $y=2$ for large positive $x$.

Answer

The limit is $2$, so $y=2$ is a horizontal asymptote.

52637712

Analyze $f(x)=\frac{x^2-1}{x^2-2x+1}$ at $x=1$. Evaluate $\lim_{x\to 1}f(x)$, or explain why it does not exist, and determine whether $x=1$ is a removable discontinuity or a vertical asymptote.

Hints

- Factor the numerator and denominator first. - Can the rational expression be simplified? - What happens when the numerator approaches a nonzero number while the denominator approaches zero? - Compare the sign of the denominator on the two sides of $x=1$.

Solution

1. Factor the numerator and denominator: $x^2-1=(x-1)(x+1)$ and $x^2-2x+1=(x-1)^2$. 2. For $x\neq 1$, simplify: $f(x)=\frac{x+1}{x-1}$. 3. As $x\to 1$, the numerator approaches $2$, while the denominator approaches $0$. 4. The left-hand limit is $\lim_{x\to 1^-}f(x)=-\infty$, and the right-hand limit is $\lim_{x\to 1^+}f(x)=+\infty$. 5. Because the one-sided limits are different, the two-sided limit does not exist. The discontinuity is infinite, and the vertical asymptote is $x=1$.

Answer

$\lim_{x\to 1}f(x)$ does not exist because $\lim_{x\to 1^-}f(x)=-\infty$ and $\lim_{x\to 1^+}f(x)=+\infty$. Therefore, $x=1$ is a vertical asymptote, not a removable discontinuity.

53005512

Consider the family of functions $f_n(x)=\frac{1}{(x-5)^n}$, where $n\in\mathbb{N}\setminus\{0\}$. 1. Suppose $n$ is odd. Find $\lim_{x\to 5^+}f_n(x)$ and $\lim_{x\to 5^-}f_n(x)$. 2. Suppose $n$ is even. Explain why $f_n(x)>0$ for every input in its domain, and find $\lim_{x\to 5}f_n(x)$.

Hints

- What sign results when a negative number is raised to an odd power? - Is $x-5$ positive or negative just to the left of $5$? - What sign does any nonzero real number have after it is raised to an even power?

Solution

1. If $n$ is odd and $x\to 5^+$, then $x-5\to 0^+$, so $(x-5)^n\to 0^+$. Therefore, $\lim_{x\to 5^+}f_n(x)=+\infty$. 2. If $n$ is odd and $x\to 5^-$, then $x-5\to 0^-$. An odd power remains negative, so $(x-5)^n\to 0^-$. Therefore, $\lim_{x\to 5^-}f_n(x)=-\infty$. 3. If $n$ is even, then $(x-5)^n>0$ for every $x\neq 5$. Since the numerator is also positive, $f_n(x)>0$ throughout its domain. 4. As $x\to 5$, the even-powered denominator approaches $0^+$ from both sides. Thus, $\lim_{x\to 5}f_n(x)=+\infty$.

Answer

1. For odd $n$, $\lim_{x\to 5^+}f_n(x)=+\infty$ and $\lim_{x\to 5^-}f_n(x)=-\infty$. 2. For even $n$, $(x-5)^n>0$ whenever $x\neq 5$, so $f_n(x)>0$. Also, $\lim_{x\to 5}f_n(x)=+\infty$.

53005612

Consider $g_k(x)=\frac{-2}{(x+1)^k}$, where $k\in\mathbb{N}\setminus\{0\}$. Find $\lim_{x\to -1^-}g_k(x)$ for odd values of $k$ and for even values of $k$. Explain how the negative numerator affects the sign of each limit.

Hints

- Determine the sign of the denominator when $x<-1$, considering whether $k$ is odd or even. - How does the numerator $-2$ affect the sign of the quotient? - Separate the sign of the denominator from the sign of the entire fraction.

Solution

1. As $x\to -1^-$, the expression $x+1$ is negative and approaches $0$. 2. If $k$ is odd, then $(x+1)^k\to 0^-$. Dividing the negative numerator $-2$ by a negative denominator gives a positive result, so $\lim_{x\to -1^-}g_k(x)=+\infty$. 3. If $k$ is even, then $(x+1)^k\to 0^+$. Dividing $-2$ by a positive denominator gives a negative result, so $\lim_{x\to -1^-}g_k(x)=-\infty$. 4. Compared with a positive numerator, the negative numerator reverses the sign of the function values.

Answer

For odd $k$, $\lim_{x\to -1^-}g_k(x)=+\infty$. For even $k$, $\lim_{x\to -1^-}g_k(x)=-\infty$. The negative numerator reverses the sign determined by the denominator.

54261912

Consider $g(x)=\frac{1}{\ln x}$. Identify its vertical asymptote and determine the one-sided behavior there. Then find its horizontal asymptote as $x\to\infty$.

Hints

- First find where the denominator can approach zero within the logarithm's domain. - The sign of the logarithm changes as its input passes through $1$. - For end behavior, ask what happens to the reciprocal when the denominator grows without bound.

Solution

1. A vertical asymptote can occur where $\ln x=0$, which is at $x=1$. 2. For $0<x<1$, $\ln x<0$ and approaches $0$ from below as $x\to1^-$, so $g(x)\to-\infty$. 3. For $x>1$, $\ln x>0$ and approaches $0$ from above as $x\to1^+$, so $g(x)\to+\infty$. Thus $x=1$ is a vertical asymptote. 4. As $x\to\infty$, $\ln x\to\infty$, so $\frac{1}{\ln x}\to0$. Therefore $y=0$ is the horizontal asymptote on the right.

Answer

Vertical asymptote: $x=1$, with $\lim_{x\to1^-}g(x)=-\infty$ and $\lim_{x\to1^+}g(x)=+\infty$. Horizontal asymptote as $x\to\infty$: $y=0$.

54262012

A function is defined by $p(x)=\frac{e^x}{1+e^x}$. Find its end behavior as $x\to\infty$ and $x\to-\infty$, and state the horizontal asymptote associated with each end.

Hints

- At the positive end, rewrite the fraction so the exponential no longer grows in both numerator and denominator. - At the negative end, use the limiting behavior of $e^x$.

Solution

1. As $x\to\infty$, divide numerator and denominator by $e^x$: $p(x)=\frac{1}{e^{-x}+1}\to1$. 2. Therefore the graph approaches $y=1$ as $x\to\infty$. 3. As $x\to-\infty$, $e^x\to0$, so $p(x)=\frac{e^x}{1+e^x}\to0$. 4. Therefore the graph approaches $y=0$ as $x\to-\infty$.

Answer

$\lim_{x\to\infty}p(x)=1$, so $y=1$ is the right-end horizontal asymptote. $\lim_{x\to-\infty}p(x)=0$, so $y=0$ is the left-end horizontal asymptote.

54262212

Consider $h(x)=\frac{\ln x}{x}$, defined for $x>0$. Analyze the graph near the left boundary of its domain and as $x\to\infty$. State any vertical or horizontal asymptotes that follow from these limits.

Hints

- Treat the boundary $x=0$ using only inputs from the function's domain. - Compare the long-run growth rates of a logarithm and a linear function.

Solution

1. As $x\to0^+$, $\ln x\to-\infty$ while $x$ remains positive and approaches $0$. Thus $\frac{\ln x}{x}\to-\infty$. 2. Therefore $x=0$ is a vertical asymptote, approached from the right. 3. As $x\to\infty$, logarithmic growth is slower than linear growth, so $\frac{\ln x}{x}\to0$. 4. Therefore $y=0$ is a horizontal asymptote as $x\to\infty$.

Answer

$\lim_{x\to0^+}h(x)=-\infty$, so $x=0$ is a vertical asymptote. $\lim_{x\to\infty}h(x)=0$, so $y=0$ is a horizontal asymptote on the right.

54262312

Let $q(x)=xe^{-x}$. Determine the end behavior of $q$ as $x\to\infty$ and as $x\to-\infty$. State whether the graph has a horizontal asymptote on either end.

Hints

- Rewrite the positive-end expression as a quotient to compare growth rates. - At the negative end, track both the sign of $x$ and the behavior of the exponential factor.

Solution

1. As $x\to\infty$, $q(x)=\frac{x}{e^x}$, and exponential growth dominates linear growth, so $q(x)\to0$. 2. Therefore $y=0$ is a horizontal asymptote as $x\to\infty$. 3. As $x\to-\infty$, the factor $x$ is negative with unbounded magnitude and $e^{-x}\to\infty$, so their product tends to $-\infty$. 4. Since the left-end limit is not finite, there is no horizontal asymptote as $x\to-\infty$.

Answer

$\lim_{x\to\infty}q(x)=0$, so $y=0$ is a right-end horizontal asymptote. $\lim_{x\to-\infty}q(x)=-\infty$, so there is no left-end horizontal asymptote.

54262612

Let $f(x)=\frac{x^3+1}{x^2+1}$. Determine whether the graph has any vertical asymptotes. Then find a slant asymptote by rewriting the function so that the difference between $f(x)$ and a line is visible.

Hints

- Check the denominator before assuming a rational function has a vertical asymptote. - Divide the polynomials and examine the leftover term at large positive and negative inputs.

Solution

1. Since $x^2+1>0$ for every real $x$, the denominator never vanishes, so there are no vertical asymptotes. 2. Polynomial division gives $f(x)=x+\frac{1-x}{x^2+1}$. 3. As $x\to\infty$ or $x\to-\infty$, $\frac{1-x}{x^2+1}\to0$. 4. Therefore $f(x)-x\to0$ on both ends, so the slant asymptote is $y=x$.

Answer

There are no vertical asymptotes. The slant asymptote is $y=x$ as $x\to\pm\infty$.

54262912

Let $f(x)=\frac{1}{\sqrt{5-x}}$, with its largest real domain. Determine the domain, describe the behavior as $x\to5^-$, and find the end behavior as $x\to-\infty$. State the asymptotes supported by these limits.

Hints

- The square root is in the denominator, so its radicand must be strictly positive. - A vertical asymptote can be established by unbounded behavior from just one side. - For the far-left behavior, track what happens to the denominator's size.

Solution

1. The denominator requires $5-x>0$, so the domain is $(-\infty, 5)$. 2. As $x\to5^-$, $\sqrt{5-x}\to0^+$, so $f(x)\to+\infty$. Thus $x=5$ is a vertical asymptote even though the function exists only on one side of it. 3. As $x\to-\infty$, $5-x\to\infty$, so $\sqrt{5-x}\to\infty$ and $f(x)\to0$. 4. Therefore $y=0$ is a horizontal asymptote on the left.

Answer

Domain: $(-\infty, 5)$. $\lim_{x\to5^-}f(x)=+\infty$, so $x=5$ is a vertical asymptote. $\lim_{x\to-\infty}f(x)=0$, so $y=0$ is a left-end horizontal asymptote.

55593112

The table compares two ratios for increasingly large positive $x$. <table><tr><th>$x$</th><td>$10$</td><td>$20$</td><td>$40$</td><td>$80$</td></tr><tr><th>$\frac{x^3}{2^x}$</th><td>$0.9766$</td><td>$0.007629$</td><td>$5.82\times10^{-8}$</td><td>$4.24\times10^{-19}$</td></tr><tr><th>$\frac{\ln x}{x}$</th><td>$0.2303$</td><td>$0.1498$</td><td>$0.0922$</td><td>$0.0548$</td></tr></table> a) What end-behavior trend does each ratio suggest? b) For each ratio, identify which part is growing faster: $2^x$ or $x^3$, and $x$ or $\ln x$. c) Based on these trends, what horizontal asymptotes would you expect for $A(x)=5+\frac{x^3}{2^x}$ and $B(x)=\frac{\ln x}{x}$ as $x\to\infty$? Explain the limitation of using a finite table as evidence.

Hints

- Interpret each displayed ratio as a comparison of the numerator's growth with the denominator's growth. - Ask what it means when a positive ratio becomes tiny as $x$ grows. - For each transformed function, connect the ratio's apparent end behavior to the constant value approached by the whole expression.

Solution

1. The values of $x^3/2^x$ rapidly decrease toward $0$, while the values of $\ln x/x$ also decrease toward $0$. 2. A ratio tending toward $0$ indicates that its denominator is outgrowing its numerator: the table therefore suggests that $2^x$ outgrows $x^3$, and $x$ outgrows $\ln x$. 3. If those end-behavior trends continue, $A(x)\to5$, so the expected horizontal asymptote is $y=5$. Also $B(x)\to0$, so the expected horizontal asymptote is $y=0$. 4. The finite table provides numerical evidence and motivation for these relative-growth facts; by itself it is not a proof of behavior for arbitrarily large $x$.

Answer

a) Both ratios suggest a limit of $0$ as $x\to\infty$. b) The data suggest that $2^x$ grows faster than $x^3$, and $x$ grows faster than $\ln x$. c) Expected asymptotes: $A$: $y=5$; $B$: $y=0$. A finite table supports these conclusions numerically but does not prove the limiting behavior for all sufficiently large $x$.

52197112

Consider the rational function $f(x)=\frac{x^2-3x}{x^2-9}$. a) State the domain of $f$ and classify each discontinuity as removable or infinite. b) Give the equation of the vertical asymptote. c) Use one-sided limits to describe the behavior of $f(x)$ near the vertical asymptote.

Hints

- First find the values that make the denominator zero. - Factor and simplify the rational expression before classifying the discontinuities. - What does a common numerator and denominator factor imply about the discontinuity? - Check the sign of the simplified expression on each side of the vertical asymptote.

Solution

1. Factor the denominator: $x^2-9=(x-3)(x+3)$. Therefore, $D_f=\mathbb{R}\setminus\{-3, 3\}$. 2. Factor the numerator and simplify for $x\neq 3$: $f(x)=\frac{x(x-3)}{(x-3)(x+3)}=\frac{x}{x+3}$. 3. The factor $x-3$ cancels, and $\lim_{x\to 3}\frac{x}{x+3}=\frac{1}{2}$. Thus, $x=3$ is a removable discontinuity. 4. The factor $x+3$ remains in the denominator, so $x=-3$ is an infinite discontinuity and the vertical asymptote is $x=-3$. 5. As $x\to -3^-$, the numerator is negative and the denominator approaches $0^-$, so $\lim_{x\to -3^-}f(x)=+\infty$. 6. As $x\to -3^+$, the numerator is negative and the denominator approaches $0^+$, so $\lim_{x\to -3^+}f(x)=-\infty$.

Answer

a) $D_f=\mathbb{R}\setminus\{-3, 3\}$. The discontinuity at $x=3$ is removable, and the discontinuity at $x=-3$ is infinite. b) $x=-3$ c) $\lim_{x\to -3^-}f(x)=+\infty$ and $\lim_{x\to -3^+}f(x)=-\infty$.

52197212

Let $h(x)=\frac{5-x}{x^2-10x+25}$. a) State the domain of $h$. b) Show algebraically that the discontinuity is a vertical asymptote. c) Find the one-sided limits of $h(x)$ as $x$ approaches $5$.

Hints

- Rewrite the denominator as a perfect-square binomial. - Pay close attention to the negative sign when rewriting the numerator. - If a denominator factor remains after complete simplification, the excluded value is a vertical asymptote. - Determine the sign of $x-5$ just to the left and just to the right of $5$.

Solution

1. Factor the denominator: $x^2-10x+25=(x-5)^2$. Therefore, the domain is $D_h=\mathbb{R}\setminus\{5\}$. 2. Rewrite the numerator as $5-x=-(x-5)$. For $x\neq 5$, $h(x)=\frac{-(x-5)}{(x-5)^2}=-\frac{1}{x-5}$. A denominator factor remains after simplification, so $x=5$ is a vertical asymptote. 3. As $x\to 5^-$, $x-5\to 0^-$, so $-\frac{1}{x-5}\to +\infty$. Thus, $\lim_{x\to 5^-}h(x)=+\infty$. 4. As $x\to 5^+$, $x-5\to 0^+$, so $-\frac{1}{x-5}\to -\infty$. Thus, $\lim_{x\to 5^+}h(x)=-\infty$.

Answer

a) $D_h=\mathbb{R}\setminus\{5\}$ b) For $x\neq 5$, $h(x)=-\frac{1}{x-5}$. Since a denominator factor remains, $x=5$ is a vertical asymptote. c) $\lim_{x\to 5^-}h(x)=+\infty$ and $\lim_{x\to 5^+}h(x)=-\infty$.

52200212

Let $g(x)=1-\frac{9}{x^2-16}$. a) State the domain of $g$. b) Find $\lim_{x\to -4^-}g(x)$, $\lim_{x\to -4^+}g(x)$, $\lim_{x\to 4^-}g(x)$, and $\lim_{x\to 4^+}g(x)$. c) Use a limit as $x\to\infty$ to determine the horizontal asymptote of the graph.

Hints

- For which values is the denominator zero? - Near each excluded value, determine whether the denominator approaches $0^+$ or $0^-$. - What happens to the fraction as $x$ becomes very large?

Solution

1. The denominator is zero when $x^2-16=0$, so $x=\pm 4$. Therefore, $D_g=\mathbb{R}\setminus\{-4, 4\}$. 2. As $x\to 4^-$, $x^2-16\to 0^-$. Thus, $\frac{9}{x^2-16}\to -\infty$, and $g(x)\to +\infty$. 3. As $x\to 4^+$, $x^2-16\to 0^+$. Thus, $\frac{9}{x^2-16}\to +\infty$, and $g(x)\to -\infty$. 4. Because $g$ is even, the behavior at $x=-4$ is reflected across the y-axis. Therefore, $\lim_{x\to -4^-}g(x)=-\infty$ and $\lim_{x\to -4^+}g(x)=+\infty$. 5. As $x\to\infty$, $\frac{9}{x^2-16}\to 0$, so $\lim_{x\to\infty}g(x)=1$. The horizontal asymptote is $y=1$.

Answer

a) $D_g=\mathbb{R}\setminus\{-4, 4\}$ b) $\lim_{x\to -4^-}g(x)=-\infty$, $\lim_{x\to -4^+}g(x)=+\infty$, $\lim_{x\to 4^-}g(x)=+\infty$, and $\lim_{x\to 4^+}g(x)=-\infty$ c) $y=1$

52617112

Determine the end behavior of $f(x)=(x^3+1)e^{-x}$ as $x\to\infty$ and as $x\to-\infty$.

Hints

- Analyze the two factors separately. - Compare the growth rate of a polynomial with the growth or decay rate of an exponential function. - Track the sign of each factor for large negative values of $x$.

Solution

1. As $x\to\infty$, the exponential decay of $e^{-x}$ is faster than the polynomial growth of $x^3+1$. Therefore, $(x^3+1)e^{-x}\to0$. 2. As $x\to-\infty$, $x^3+1\to-\infty$ and $e^{-x}\to\infty$. The product is negative with unbounded magnitude, so $f(x)\to-\infty$.

Answer

$\lim_{x\to\infty}f(x)=0$ and $\lim_{x\to-\infty}f(x)=-\infty$

52617212

Determine the end behavior of $g(x)=2x^2-e^{0.1x}$ as $x\to\infty$ and as $x\to-\infty$.

Hints

- Identify which term has the faster growth rate for large positive $x$. - Factoring out the exponential term can make the dominant behavior clearer. - Determine what $e^{0.1x}$ does as $x\to-\infty$.

Solution

1. As $x\to\infty$, the exponential term grows faster than the polynomial term. Write $g(x)=e^{0.1x}\left(2x^2e^{-0.1x}-1\right)$. Since $2x^2e^{-0.1x}\to0$, the expression in parentheses approaches $-1$, so $g(x)\to-\infty$. 2. As $x\to-\infty$, $2x^2\to\infty$ while $e^{0.1x}\to0$. Therefore, $g(x)\to\infty$.

Answer

$\lim_{x\to\infty}g(x)=-\infty$ and $\lim_{x\to-\infty}g(x)=\infty$

52618912

Let $f(x)=\frac{100x^2+e^x}{e^x-x^4}$. Find $\lim_{x\to\infty}f(x)$ and $\lim_{x\to-\infty}f(x)$. Justify each result by comparing the growth rates of the terms.

Hints

- For large positive $x$, divide by the fastest-growing term. - Recall that $e^x$ grows faster than any power of $x$. - For large negative $x$, determine what happens to $e^x$. - Analyze the two directions separately.

Solution

1. As $x\to\infty$, divide the numerator and denominator by $e^x$: $f(x)=\frac{100x^2/e^x+1}{1-x^4/e^x}$. Since every polynomial grows more slowly than $e^x$, both polynomial-to-exponential ratios approach $0$. Therefore, $\lim_{x\to\infty}f(x)=1$. 2. As $x\to-\infty$, $e^x\to0$, so the polynomial terms dominate. Dividing the numerator and denominator by $x^4$ gives $f(x)=\frac{100/x^2+e^x/x^4}{e^x/x^4-1}$. The numerator approaches $0$ and the denominator approaches $-1$, so $\lim_{x\to-\infty}f(x)=0$.

Answer

$\lim_{x\to\infty}f(x)=1$ and $\lim_{x\to-\infty}f(x)=0$

52619012

Let $g(x)=(x^2-5x)e^x$. A student claims, “As $x\to-\infty$, the first factor approaches $+\infty$ and the second factor approaches $0$, so they cancel and the limit must be $1$.” Evaluate the student's claim and find the correct limit as $x\to-\infty$.

Hints

- A product with one factor approaching infinity and the other approaching zero does not have an automatic value. - Rewrite $e^x$ using $u=-x$. - Compare polynomial growth with exponential growth.

Solution

1. The claim is incorrect. A product of a quantity approaching infinity and a quantity approaching zero is an indeterminate form; the limit depends on their relative rates. 2. Let $u=-x$. As $x\to-\infty$, $u\to\infty$, and $g(x)=(u^2+5u)e^{-u}=\frac{u^2+5u}{e^u}$. 3. The exponential denominator grows faster than the polynomial numerator, so $\frac{u^2+5u}{e^u}\to0$. Therefore, $\lim_{x\to-\infty}g(x)=0$.

Answer

The claim is incorrect; $\lim_{x\to-\infty}(x^2-5x)e^x=0$.

52635612

Let $f(x)=\frac{2x+4}{x-1}$. a) Find $\lim_{x\to\infty}f(x)$. b) Find $\lim_{x\to1^+}f(x)$ and $\lim_{x\to1^-}f(x)$. c) State the horizontal and vertical asymptotes.

Hints

- Divide by $x$ for the limit at infinity. - Determine the sign of the denominator on each side of $x=1$. - Use the limits to identify the asymptotes.

Solution

1. Divide the numerator and denominator by $x$: $f(x)=\frac{2+\frac{4}{x}}{1-\frac{1}{x}}$. Therefore, $f(x)\to2$ as $x\to\infty$. 2. As $x\to1^+$, the numerator approaches $6$ and the denominator approaches $0$ through positive values, so $f(x)\to+\infty$. 3. As $x\to1^-$, the numerator approaches $6$ and the denominator approaches $0$ through negative values, so $f(x)\to-\infty$. 4. Thus, the horizontal asymptote is $y=2$, and the vertical asymptote is $x=1$.

Answer

a) $2$ b) $\lim_{x\to1^+}f(x)=+\infty$; $\lim_{x\to1^-}f(x)=-\infty$ c) Horizontal: $y=2$; vertical: $x=1$

52739912

Evaluate each limit. a) $\lim_{x\to\infty}\frac{x^2+4}{e^x}$ b) $\lim_{x\to-\infty}(x-2)e^x$ c) $\lim_{x\to0^+}\frac{3}{1-e^x}$

Hints

- Compare polynomial and exponential growth rates. - Determine what $e^x$ does as $x\to-\infty$. - For a denominator approaching zero, determine whether it approaches from the positive or negative side.

Solution

1. In part a, $e^x$ grows faster than the polynomial $x^2+4$, so the quotient approaches $0$. 2. In part b, let $u=-x$. Then $(x-2)e^x=(-u-2)e^{-u}=-\frac{u+2}{e^u}$, which approaches $0$ as $u\to\infty$. 3. In part c, as $x\to0^+$, $e^x>1$, so $1-e^x\to0^-$. Therefore, $\frac{3}{1-e^x}\to-\infty$.

Answer

a) $0$ b) $0$ c) $-\infty$

52740112

Let $f(x)=\frac{x+2}{e^x-e^2}$. Determine the domain and analyze the behavior of $f$ at every boundary of its domain.

Hints

- Find where the denominator is zero. - Compare linear and exponential growth for large positive $x$. - Determine the limiting denominator as $x\to-\infty$. - Analyze the signs on the two sides of the excluded value.

Solution

1. The denominator is zero when $e^x=e^2$, so $x=2$. Thus, the domain is $D=\mathbb{R}\setminus\{2\}$. 2. As $x\to\infty$, the exponential denominator grows faster than the linear numerator, so $f(x)\to0$. 3. As $x\to-\infty$, $e^x\to0$, so the denominator approaches $-e^2$, while the numerator approaches $-\infty$. Therefore, $f(x)\to\infty$. 4. As $x\to2^+$, the numerator approaches $4$ and the denominator approaches $0^+$, so $f(x)\to\infty$. As $x\to2^-$, the denominator approaches $0^-$, so $f(x)\to-\infty$.

Answer

$D=\mathbb{R}\setminus\{2\}$ $\lim_{x\to\infty}f(x)=0$ $\lim_{x\to-\infty}f(x)=\infty$ $\lim_{x\to2^+}f(x)=\infty$ $\lim_{x\to2^-}f(x)=-\infty$

52751112

Let $f(x)=4-\sqrt{2x+6}$, with largest real domain $D_f$. a) Find $D_f$ and the range of $f$. b) Find the equation of the tangent line $t$ to the graph of $f$ at $x=\frac{3}{2}$. c) Find $\lim_{x\to-3^+}f'(x)$, and describe its geometric meaning for the graph of $f$.

Hints

- Require the expression under the square root to be nonnegative. - Use the sign before the radical to determine the range. - Apply the chain rule to find the derivative. - Use the point and derivative value to write the tangent line. - Interpret an unbounded derivative near an endpoint geometrically.

Solution

1. a) Require $2x+6\ge0$, so $D_f=[-3,\infty)$. Since $\sqrt{2x+6}\ge0$ and grows without bound, the range is $(-\infty,4]$. 2. b) Differentiate: $f'(x)=-\frac{1}{\sqrt{2x+6}}$. 3. At $x=\frac{3}{2}$, $f\left(\frac{3}{2}\right)=1$ and $f'\left(\frac{3}{2}\right)=-\frac{1}{3}$. 4. Using point-slope form, $y-1=-\frac{1}{3}\left(x-\frac{3}{2}\right)$, so $t:y=-\frac{1}{3}x+\frac{3}{2}$. 5. c) As $x\to-3^+$, $\sqrt{2x+6}\to0^+$, so $f'(x)\to-\infty$. The graph has a vertical tangent at its endpoint $(-3,4)$.

Answer

a) $D_f=[-3,\infty)$; range: $(-\infty,4]$ b) $t:y=-\frac{1}{3}x+\frac{3}{2}$ c) $\lim_{x\to-3^+}f'(x)=-\infty$; the graph has a vertical tangent at $(-3,4)$.

52752912

Let $f(x)=2\sqrt{9-x^2}$, with largest real domain $D_f$. a) Find $D_f$, and determine whether the graph of $f$ has symmetry about either coordinate axis. b) Find $f'(x)$. Determine the behavior of $f'(x)$ as $x\to3^-$ and as $x\to-3^+$. What is the geometric position of the tangent lines at the endpoints of the domain?

Hints

- Require the square-root radicand to be nonnegative. - Replace $x$ with $-x$ to test symmetry. - Apply the chain rule to differentiate. - Analyze the signs of the numerator and denominator near each endpoint. - An unbounded derivative magnitude indicates a vertical tangent.

Solution

1. a) Require $9-x^2\ge0$, so $D_f=[-3,3]$. Since $f(-x)=f(x)$, the graph is symmetric about the y-axis. It is not symmetric about the x-axis because it contains only nonnegative y-values. 2. b) By the chain rule, $f'(x)=-\frac{2x}{\sqrt{9-x^2}}$. 3. As $x\to3^-$, the numerator approaches $-6$ and the denominator approaches $0^+$, so $f'(x)\to-\infty$. 4. As $x\to-3^+$, the numerator approaches $6$ and the denominator approaches $0^+$, so $f'(x)\to\infty$. 5. Therefore, the graph has vertical tangent lines at $(-3,0)$ and $(3,0)$.

Answer

a) $D_f=[-3,3]$; the graph is symmetric about the y-axis and not symmetric about the x-axis. b) $f'(x)=-\frac{2x}{\sqrt{9-x^2}}$; $f'(x)\to-\infty$ as $x\to3^-$, and $f'(x)\to\infty$ as $x\to-3^+$. Both endpoint tangents are vertical.

52753012

Let $g(x)=\sqrt{100-4x^2}$ on its largest real domain. a) Find the domain of $g$ and the x-intercepts of its graph. b) Find the equation of the tangent line to the graph of $g$ at $x_0=3$. c) Analyze the slope near the endpoints of the domain. Use this behavior to explain why $g$ is not differentiable at $x=-5$ or $x=5$.

Hints

- Set the radicand greater than or equal to $0$ to find the domain. - Use the derivative and graph point to write the tangent line. - Examine the radical denominator as $x$ approaches each endpoint. - A derivative that becomes unbounded does not have a finite endpoint value.

Solution

1. a) Require $100-4x^2\ge0$, so $x^2\le25$ and the domain is $[-5,5]$. The x-intercepts are $(-5,0)$ and $(5,0)$. 2. Differentiate: $g'(x)=-\frac{4x}{\sqrt{100-4x^2}}$. 3. b) At $x=3$, $g(3)=8$ and $g'(3)=-\frac{3}{2}$. Thus, $y-8=-\frac{3}{2}(x-3)$, or $y=-\frac{3}{2}x+\frac{25}{2}$. 4. c) As $x\to5^-$, $g'(x)\to-\infty$. As $x\to-5^+$, $g'(x)\to\infty$. 5. The endpoint slopes do not approach finite real values, so the function is not differentiable at the endpoints; the graph has vertical tangents there.

Answer

a) Domain $[-5,5]$; x-intercepts $(-5,0)$ and $(5,0)$ b) $y=-\frac{3}{2}x+\frac{25}{2}$ c) The slope approaches $-\infty$ at $5$ from the left and $\infty$ at $-5$ from the right, so $g$ is not differentiable at either endpoint.

52762512

Analyze the behavior of $f(x)=(x^2+3x)\ln(x)$ as $x\to0^+$. Justify the result by comparing the rates at which power and logarithmic functions approach their limits.

Hints

- Track the behavior of both factors as $x\to0^+$. - Expand the polynomial factor into two terms. - Recall the standard limit $x^p\ln(x)\to0$ for $p>0$. - Apply the sum law for limits.

Solution

1. Expand the product: $f(x)=x^2\ln(x)+3x\ln(x)$. 2. For every positive exponent $p$, $x^p\ln(x)\to0$ as $x\to0^+$. The power factor approaches $0$ rapidly enough to dominate the logarithm's divergence to $-\infty$. 3. Therefore, $x^2\ln(x)\to0$ and $3x\ln(x)\to0$. 4. Hence, $\lim_{x\to0^+}(x^2+3x)\ln(x)=0$.

Answer

$\lim_{x\to0^+}(x^2+3x)\ln(x)=0$

52762612

Let $g(x)=\frac{2x^2-\ln(x)}{x^2+5}$. Determine the behavior of $g(x)$ as $x\to\infty$.

Hints

- Divide the numerator and denominator by the highest power of $x$. - Compare the growth of $x^2$ with the growth of $\ln(x)$. - Determine the limits of the smaller fractions separately. - Apply the quotient law for limits.

Solution

1. Divide the numerator and denominator by $x^2$: $g(x)=\frac{2-\frac{\ln(x)}{x^2}}{1+\frac{5}{x^2}}$. 2. As $x\to\infty$, $\frac{5}{x^2}\to0$. 3. A power function grows faster than a logarithmic function, so $\frac{\ln(x)}{x^2}\to0$. 4. Therefore, $\lim_{x\to\infty}g(x)=\frac{2-0}{1+0}=2$.

Answer

$g(x)\to2$ as $x\to\infty$.

52764112

Let $f(x)=\frac{\ln(x^2)+1}{x^2}$, where $x\neq0$. Determine the behavior of $f(x)$ as $x\to0$ and as $x\to\infty$.

Hints

- Examine the numerator and denominator separately as $x\to0$. - Compare the growth of a logarithmic function with the growth of a power function. - Split the expression into two fractions for the limit at infinity.

Solution

1. As $x\to0$, $x^2\to0^+$ and $\ln(x^2)+1\to-\infty$. Therefore, $\frac{\ln(x^2)+1}{x^2}\to-\infty$. 2. As $x\to\infty$, rewrite the function as $f(x)=\frac{\ln(x^2)}{x^2}+\frac{1}{x^2}$. A power function grows faster than a logarithmic function, so $\frac{\ln(x^2)}{x^2}\to0$, and $\frac{1}{x^2}\to0$. 3. Thus, $\lim_{x\to0}f(x)=-\infty$ and $\lim_{x\to\infty}f(x)=0$.

Answer

As $x\to0$, $f(x)\to-\infty$. As $x\to\infty$, $f(x)\to0$.

53005212

Let $f(x)=\frac{1}{2}x^2+3x+4e^{-x/2}$. a) Show that the graph of $f$ approaches a parabola $q$ as $x\to\infty$, and give the equation of $q$. b) Determine whether the graph also approaches $q$ as $x\to-\infty$. Justify your answer with a limit.

Hints

- Identify the polynomial and exponential parts of the function. - Two graphs approach each other when the difference of their function values approaches $0$. - Review the behavior of $e^u$ as $u\to\infty$ and $u\to-\infty$. - Evaluate the two directions separately.

Solution

1. Separate the polynomial and exponential parts: $q(x)=\frac{1}{2}x^2+3x$ and $f(x)-q(x)=4e^{-x/2}$. 2. As $x\to\infty$, $\lim(f(x)-q(x))=\lim4e^{-x/2}=0$. Therefore, the graph of $f$ approaches the parabola $q(x)=\frac{1}{2}x^2+3x$. 3. As $x\to-\infty$, the exponent $-x/2\to\infty$, so $\lim(f(x)-q(x))=\lim4e^{-x/2}=\infty$. Therefore, the graph does not approach $q$ in that direction.

Answer

a) $q(x)=\frac{1}{2}x^2+3x$, because $\lim_{x\to\infty}(f(x)-q(x))=0$. b) No. Since $\lim_{x\to-\infty}(f(x)-q(x))=\infty$, the graphs do not approach each other as $x\to-\infty$.

53264312

Panel a) shows the graph of $f'$ on $-2.5\le x\le2.5$. Panels b), c), and d) are candidate graphs for $f$ on the same displayed interval. Which candidate graph is consistent with $f'$? Justify your choice from the sign of $f'$. Then state the displayed intervals on which $f$ is increasing or decreasing and give the x-coordinates of the local extrema indicated by the derivative sign changes.

Hints

- Read where the derivative graph is above and below the x-axis. - Translate positive derivative into increasing behavior and negative derivative into decreasing behavior. - Look for sign changes of the derivative before comparing the candidate function graphs.

Solution

1. In panel a), $f'(x)>0$ for $-2.5<x<-1$, $f'(x)<0$ for $-1<x<1$, and $f'(x)>0$ for $1<x<2.5$. 2. Therefore $f$ must increase, then decrease, then increase across those three intervals. 3. Candidate b) has exactly that monotonicity pattern. Candidate c) has the opposite pattern, and candidate d) is increasing throughout the displayed interval. 4. Because $f'$ changes from positive to negative at $x=-1$, $f$ has a local maximum there. Because $f'$ changes from negative to positive at $x=1$, $f$ has a local minimum there.

Answer

Panel b) is consistent with $f'$. $f$ is increasing on $(-2.5,-1)$, decreasing on $(-1,1)$, and increasing on $(1,2.5)$. Local maximum x-coordinate: $-1$. Local minimum x-coordinate: $1$.

53452512

Let $h(x)=\frac{e^x}{x+2}$, with domain $\mathbb R\setminus\{-2\}$. a) Explain why $h$ has no zeros. b) Analyze the behavior of $h$ near $x=-2$, as $x\to\infty$, and as $x\to-\infty$. c) Find the coordinates of the local extremum of the graph.

Hints

- A quotient is zero only when its numerator is zero. - Track the sign of the denominator on each side of $x=-2$. - Compare exponential and linear growth for large positive $x$. - Use the quotient rule to locate the turning point after the end behavior is established.

Solution

1. A quotient is zero only when its numerator is zero. Since $e^x>0$ for every real $x$, $h$ has no zeros. 2. Near $x=-2$, the numerator approaches $e^{-2}>0$. The denominator approaches $0^+$ from the right and $0^-$ from the left, so $\lim_{x\to-2^+}h(x)=\infty$ and $\lim_{x\to-2^-}h(x)=-\infty$. As $x\to\infty$, exponential growth dominates the linear denominator, so $h(x)\to\infty$. As $x\to-\infty$, $e^x\to0$, so $h(x)\to0$. 3. By the quotient rule, $h'(x)=\frac{e^x(x+1)}{(x+2)^2}$. The only critical number is $x=-1$. The derivative changes from negative to positive, so the graph has a local minimum at $\left(-1,\frac{1}{e}\right)$.

Answer

a) No zeros b) $\lim_{x\to-2^+}h(x)=\infty$, $\lim_{x\to-2^-}h(x)=-\infty$, $\lim_{x\to\infty}h(x)=\infty$, and $\lim_{x\to-\infty}h(x)=0$ c) Local minimum at $\left(-1,\frac{1}{e}\right)$

53453712

Let $f(x)=\frac{1}{2}x+\frac{2}{x}$ on the closed interval $[1,6]$. Find all critical points of $f$ in $(1,6)$. Then use the candidates test to determine the absolute minimum value and absolute maximum value of $f$ on $[1,6]$, including where each occurs.

Hints

- On a closed interval, the candidates test includes interior critical points and both endpoints. - Find where the first derivative is zero inside the interval. - Compare the function values at every candidate rather than using derivative sign alone.

Solution

1. Differentiate: $f'(x)=\frac{1}{2}-\frac{2}{x^2}$. 2. A critical point occurs when $f'(x)=0$. Solving $\frac{1}{2}-\frac{2}{x^2}=0$ gives $x^2=4$, so the only critical point in $(1,6)$ is $x=2$. 3. Evaluate the candidates: $f(1)=\frac{5}{2}$, $f(2)=2$, and $f(6)=\frac{10}{3}$. 4. The smallest candidate value is $2$, so the absolute minimum is $2$ at $x=2$. The largest is $\frac{10}{3}$, so the absolute maximum is $\frac{10}{3}$ at $x=6$.

Answer

Critical point: $x=2$. Absolute minimum: $2$ at $x=2$. Absolute maximum: $\frac{10}{3}$ at $x=6$.

54261412

Let $f(x)=\frac{\sqrt{x^2+1}}{x}$, $x\ne0$. Find the one-sided limits as $x\to0$, the limits as $x\to\infty$ and $x\to-\infty$, and state all vertical and horizontal asymptotes.

Hints

- Near $0$, the numerator stays positive while the denominator changes sign. - At large magnitude, factor $x^2$ from inside the square root carefully. - Remember that $\sqrt{x^2}=|x|$, which behaves differently for positive and negative $x$.

Solution

1. As $x\to0$, the numerator approaches $1$. The denominator approaches $0^-$ from the left and $0^+$ from the right, so $\lim_{x\to0^-}f(x)=-\infty$ and $\lim_{x\to0^+}f(x)=+\infty$. Thus $x=0$ is a vertical asymptote. 2. For $x>0$, $f(x)=\sqrt{1+\frac{1}{x^2}}$, so $\lim_{x\to\infty}f(x)=1$. 3. For $x<0$, $\sqrt{x^2}=|x|=-x$, so $f(x)=-\sqrt{1+\frac{1}{x^2}}$. Hence $\lim_{x\to-\infty}f(x)=-1$. The horizontal asymptotes are $y=1$ to the right and $y=-1$ to the left.

Answer

$\lim_{x\to0^-}f(x)=-\infty$, $\lim_{x\to0^+}f(x)=+\infty$. $\lim_{x\to\infty}f(x)=1$, $\lim_{x\to-\infty}f(x)=-1$. Vertical asymptote: $x=0$. Horizontal asymptotes: $y=1$ as $x\to\infty$, and $y=-1$ as $x\to-\infty$.

54261512

Let $g(x)=\frac{1}{1-e^{-x}}$, $x\ne0$. Determine the behavior as $x\to0^-$, $x\to0^+$, $x\to\infty$, and $x\to-\infty$. State the vertical asymptote and any horizontal asymptotes.

Hints

- Near $0$, determine whether $e^{-x}$ is slightly greater than or less than $1$ on each side. - At positive infinity, track the exponential decay of $e^{-x}$. - At negative infinity, recognize that $e^{-x}$ grows without bound.

Solution

1. As $x\to0^-$, $e^{-x}>1$ and approaches $1$, so $1-e^{-x}\to0^-$ and $g(x)\to-\infty$. As $x\to0^+$, the denominator approaches $0^+$, so $g(x)\to+\infty$. Thus $x=0$ is a vertical asymptote. 2. As $x\to\infty$, $e^{-x}\to0$, so $g(x)\to1$. 3. As $x\to-\infty$, $e^{-x}\to+\infty$, so the denominator tends to $-\infty$ and $g(x)\to0$. Therefore the right-end horizontal asymptote is $y=1$, while the left-end horizontal asymptote is $y=0$.

Answer

$\lim_{x\to0^-}g(x)=-\infty$, $\lim_{x\to0^+}g(x)=+\infty$. $\lim_{x\to\infty}g(x)=1$, $\lim_{x\to-\infty}g(x)=0$. Vertical asymptote: $x=0$. Horizontal asymptotes: $y=1$ as $x\to\infty$, and $y=0$ as $x\to-\infty$.

54261612

Let $f(x)=\frac{x^2+1}{x-1}$. Find the vertical asymptote and the slant asymptote. Justify the slant asymptote with a limit involving the difference between $f(x)$ and a line.

Hints

- Divide the numerator by the denominator to separate a line from a remainder term. - An asymptotic line is confirmed when the difference between the function and the line approaches zero. - Check excluded denominator zeros separately for vertical behavior.

Solution

1. Polynomial division gives $f(x)=x+1+\frac{2}{x-1}$. 2. As $x\to\pm\infty$, $\frac{2}{x-1}\to0$, so $f(x)-(x+1)\to0$. Thus the slant asymptote is $y=x+1$. 3. At $x=1$, the numerator approaches $2$ while the denominator approaches $0$, giving unbounded one-sided behavior. Thus $x=1$ is a vertical asymptote.

Answer

Vertical asymptote: $x=1$. Slant asymptote: $y=x+1$.

54262112

Let $f(x)=\frac{3x}{\sqrt{x^2+9}}$. Find the horizontal asymptote approached as $x\to\infty$ and the horizontal asymptote approached as $x\to-\infty$. Explain why the two limits have opposite signs.

Hints

- Factor the largest power of $x$ from inside the square root. - Remember that $\sqrt{x^2}=|x|$, especially when $x$ is negative.

Solution

1. Write $\sqrt{x^2+9}=|x|\sqrt{1+9/x^2}$. 2. As $x\to\infty$, $|x|=x$, so $f(x)=\frac{3}{\sqrt{1+9/x^2}}\to3$. 3. As $x\to-\infty$, $|x|=-x$, so $f(x)=\frac{-3}{\sqrt{1+9/x^2}}\to-3$. 4. The sign difference comes from $x/|x|$, which approaches $1$ on the right and $-1$ on the left.

Answer

As $x\to\infty$, $f(x)\to3$, so $y=3$ is the right-end horizontal asymptote. As $x\to-\infty$, $f(x)\to-3$, so $y=-3$ is the left-end horizontal asymptote.

54262412

A student says, “Once a graph has a horizontal asymptote, it can never cross that line.” Consider $f(x)=1+\frac{\sin x}{x}$, for $x\ne0$. Determine the horizontal asymptote as $x\to\pm\infty$, and use this function to evaluate the student's claim.

Hints

- Separate the question of long-run behavior from what can happen at finite x-values. - Look for inputs where the added fraction becomes exactly zero.

Solution

1. Since $-1\le\sin x\le1$, the quotient $\frac{\sin x}{x}\to0$ as $x\to\infty$ and as $x\to-\infty$. 2. Therefore $f(x)\to1$ on both ends, so $y=1$ is a horizontal asymptote. 3. Whenever $x=n\pi$ for a nonzero integer $n$, $\sin x=0$, so $f(n\pi)=1$. 4. Thus the graph meets the horizontal asymptote infinitely many times. A horizontal asymptote describes end behavior; it does not forbid crossings at finite inputs.

Answer

The horizontal asymptote is $y=1$ as $x\to\pm\infty$. The student's claim is false because $f(n\pi)=1$ for every nonzero integer $n$, so the graph crosses or meets its horizontal asymptote repeatedly.

54262512

Use the graph of $s$ on $[0,2\pi]$ to locate the vertical asymptotes. Describe the one-sided behavior at each asymptote.

Hints

- Look for labeled x-values where neighboring branches leave the visible vertical scale. - Read the direction of each branch separately from the left and from the right. - Match each unbounded branch pair to the exact $\pi$-based tick label beneath it.

Solution

1. The graph becomes unbounded at $x=\frac{\pi}{2}$ and $x=\frac{3\pi}{2}$, so these are the vertical asymptotes. 2. At $x=\frac{\pi}{2}$, the left branch increases without bound and the right branch decreases without bound. Thus the left-hand limit is $+\infty$ and the right-hand limit is $-\infty$. 3. At $x=\frac{3\pi}{2}$, the left branch decreases without bound and the right branch increases without bound. Thus the left-hand limit is $-\infty$ and the right-hand limit is $+\infty$.

Answer

Vertical asymptotes: $x=\frac{\pi}{2}$ and $x=\frac{3\pi}{2}$. At $x=\frac{\pi}{2}$: left limit $+\infty$, right limit $-\infty$. At $x=\frac{3\pi}{2}$: left limit $-\infty$, right limit $+\infty$.

54262712

Consider $r(x)=\sqrt{x^2+9}$. The graph has no vertical asymptotes, but it approaches a different line at each end. Find the line approached as $x\to\infty$ and the line approached as $x\to-\infty$.

Hints

- To show a line is an asymptote, examine the difference between the function and that line. - The square root behaves like $|x|$ for large $|x|$, so the two ends should be treated separately.

Solution

1. For $x\to\infty$, compare $r(x)$ with $x$. Rationalizing gives $r(x)-x=\frac{9}{\sqrt{x^2+9}+x}\to0$. Thus the right-end slant asymptote is $y=x$. 2. For $x\to-\infty$, compare $r(x)$ with $-x$. Then $r(x)-(-x)=\sqrt{x^2+9}+x$. 3. Rationalizing this difference gives $\sqrt{x^2+9}+x=\frac{9}{\sqrt{x^2+9}-x}\to0$ as $x\to-\infty$. 4. Therefore the left-end slant asymptote is $y=-x$.

Answer

As $x\to\infty$, the graph approaches $y=x$. As $x\to-\infty$, the graph approaches $y=-x$.

54263012

Analyze $F(x)=\frac{e^x}{e^x-2}$. Find its vertical asymptote with the two one-sided infinite limits, and find the horizontal asymptote approached at each end of the x-axis.

Hints

- Solve for the input where the denominator vanishes, then check the denominator's sign on each side. - For positive-end behavior, divide through by the dominant exponential term. - For negative-end behavior, use the fact that $e^x$ approaches zero.

Solution

1. The denominator is zero when $e^x=2$, so the possible vertical asymptote is $x=\ln2$. 2. As $x\to(\ln2)^-$, the denominator approaches $0^-$ while the numerator approaches $2$, so $F(x)\to-\infty$. From the right, the denominator approaches $0^+$, so $F(x)\to+\infty$. 3. As $x\to\infty$, divide numerator and denominator by $e^x$: $F(x)=\frac{1}{1-2e^{-x}}\to1$. 4. As $x\to-\infty$, $e^x\to0$, so $F(x)\to0$.

Answer

Vertical asymptote: $x=\ln2$, with left limit $-\infty$ and right limit $+\infty$. Horizontal asymptotes: $y=1$ as $x\to\infty$ and $y=0$ as $x\to-\infty$.

54263112

Let $f(x)=e^{1/x}$, for $x\ne0$. Find the two one-sided limits as $x\to0$ and the limits as $x\to\pm\infty$. Use them to identify every vertical and horizontal asymptote.

Hints

- First determine what $1/x$ does from each side of zero. - A vertical asymptote does not require both one-sided limits to be infinite. - At very large $|x|$, the exponent itself approaches a simple value.

Solution

1. As $x\to0^+$, $1/x\to+\infty$, so $e^{1/x}\to+\infty$. This is enough to make $x=0$ a vertical asymptote. 2. As $x\to0^-$, $1/x\to-\infty$, so $e^{1/x}\to0$. The two sides therefore have very different behavior. 3. As $x\to\infty$ or $x\to-\infty$, $1/x\to0$, so $e^{1/x}\to e^0=1$. 4. Thus $y=1$ is a horizontal asymptote on both ends.

Answer

$\lim_{x\to0^+}f(x)=+\infty$ and $\lim_{x\to0^-}f(x)=0$, so $x=0$ is a vertical asymptote. $\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=1$, so $y=1$ is the horizontal asymptote on both ends.

54261712

Consider $f(x)=\sqrt{x^2+x}-x$ on its real domain. Find $\lim_{x\to\infty}f(x)$ by rationalizing, and state the corresponding horizontal asymptote. Also describe what happens as $x\to-\infty$.

Hints

- A difference of two large quantities suggests multiplying by a conjugate. - Treat positive and negative large x-values separately because $\sqrt{x^2}=|x|$. - A finite end limit gives a horizontal asymptote on that end only.

Solution

1. For $x\ne0$, rationalize: $\sqrt{x^2+x}-x=\frac{x}{\sqrt{x^2+x}+x}$. This identity is sufficient for both end limits. 2. For $x\to\infty$, divide numerator and denominator by $x>0$ to get $\frac{1}{\sqrt{1+1/x}+1}\to\frac{1}{2}$. Thus $y=\frac{1}{2}$ is a right-end horizontal asymptote. 3. As $x\to-\infty$, $\sqrt{x^2+x}\sim -x$, so $\sqrt{x^2+x}-x\sim-2x\to+\infty$. There is no horizontal asymptote on the left.

Answer

$\lim_{x\to\infty}f(x)=\frac{1}{2}$, so the right-end horizontal asymptote is $y=\frac{1}{2}$. As $x\to-\infty$, $f(x)\to+\infty$.

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