Free math worksheets

- Check that each expression is defined at the approach value. - Then substitute directly and simplify carefully.

1. Substitute \(x=2\): \(\frac{3\cdot 2^2-4}{2+2}=\frac{8}{4}=2\). 2. Substitute \(x=9\): \(\sqrt{9}\cdot(9-7)=3\cdot 2=6\). 3. Substitute \(x=-1\): \((-1)^5+3\cdot(-1)^2-1=-1+3-1=1\).

1) \(2\) 2) \(6\) 3) \(1\)

- Identify the limiting input to the outer absolute-value function. - Recall whether absolute value is continuous at negative numbers. - Apply the outer function after the inner limit is known.

1. The inner function satisfies \(p(x)\to-5\) as \(x\to-2\). 2. The absolute-value function is continuous at every real number. 3. Therefore \(\lim_{x\to-2}|p(x)|=|-5|=5\).

\(5\). Absolute value maps the limiting value \(-5\) to its nonnegative magnitude.

- Check the limiting value of the denominator before taking a reciprocal. - Recall the nonzero-denominator condition for quotient limits. - Once that condition is satisfied, take the reciprocal of the limiting value.

1. The denominator approaches \(-2\). 2. Because \(-2\ne0\), the reciprocal case of the quotient law applies. 3. Therefore \(\lim_{t\to3}\frac{1}{r(t)}=\frac{1}{-2}=-\frac{1}{2}\).

\(-\frac{1}{2}\). The law applies because \(r(t)\) approaches a nonzero value.

- Substitute \(x=-2\) into the absolute-value expression and the polynomial separately. - Evaluate the denominator before applying the quotient law. - Confirm that the denominator's limiting value is not zero.

1. The numerator approaches \(|-2-1|=3\) because absolute value is continuous. 2. The denominator approaches \((-2)^2+1=5\). 3. Since the denominator limit is nonzero, the quotient law gives \(\frac{3}{5}\).

\(\frac{3}{5}\).

- Separate the variable part from the constant part. - Recall how a fixed multiplier passes through a limit. - Combine the resulting component limits at the end.

1. By the constant multiple law, \(\lim_{x\to a}m f(x)=mL\). 2. The constant function \(b\) has limit \(b\). 3. By the sum law, \(\lim_{x\to a}(m f(x)+b)=mL+b\).

\(mL+b\).

- Substitute \(x=3\) into the numerator and denominator separately. - Distinguish a zero numerator from a zero denominator. - Apply the quotient law after checking that the denominator limit is nonzero.

1. Direct substitution gives the numerator limit \(3^2-9=0\). 2. The denominator limit is \(3+3=6\), which is nonzero. 3. Therefore the quotient law applies and gives \(\frac{0}{6}=0\).

\(0\). A zero numerator is allowed; the quotient law only requires the denominator limit to be nonzero.

- Apply the power law to each function separately. - Keep the exponents symbolic. - Combine the two resulting limits with the product law.

1. By the power law, \(\lim_{x\to a}[f(x)]^m=F^m\). 2. Similarly, \(\lim_{x\to a}[g(x)]^n=G^n\). 3. By the product law, the required limit is \(F^mG^n\).

\(F^mG^n\).

- Identify the limiting input to the outer root function. - Check the real-domain condition for an even root. - Apply the root only after confirming that condition.

1. The inner function approaches \(16\). 2. The fourth-root function is continuous for nonnegative real inputs, and \(16>0\). 3. Therefore \(\lim_{x\to1}\sqrt[4]{p(x)}=\sqrt[4]{16}=2\).

\(2\). The limiting input \(16\) is positive, so the real fourth root is defined and continuous there.

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

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 \(h(x)\to+\infty\).

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) \(\lim_{x\to\infty}h(x)=-2\); \(\lim_{x\to-\infty}h(x)=+\infty\)

- Treat the outer addition, subtraction, and powers one operation at a time. - Replace each basic function by its known limiting value only after identifying the applicable law. - Check that squaring affects the sign of the second limiting value.

1. By the constant multiple and difference laws, the limit is \(2\lim_{x\to4}f(x)-\lim_{x\to4}[g(x)]^2\). 2. By the power law, \(\lim_{x\to4}[g(x)]^2=(-2)^2=4\). 3. Therefore the limit is \(2\cdot 3-4=2\).

\(2\), using the constant multiple, difference, and power laws for limits.

- Find the numerator and denominator limits separately. - Before using a quotient law, inspect the limiting value of the denominator. - Combine the known limits only after checking that condition.

1. The numerator limit is \(5+2=7\) by the sum law. 2. The denominator limit is \(3-2=1\) by the difference law. 3. Because the denominator limit is nonzero, the quotient law applies, giving \(\frac{7}{1}=7\).

\(7\). The quotient law is valid because the denominator approaches \(1\), not \(0\).

- Work outward from the limits of \(p\) and \(q\). - Determine the limiting value of the denominator before taking a quotient. - Keep the result symbolic rather than substituting numerical values.

1. The numerator limit is \(P^2+3\) by the power and sum laws. 2. The denominator limit is \(Q-1\) by the difference law. 3. Since \(Q\ne1\), the denominator limit is nonzero, so the quotient law gives \(\frac{P^2+3}{Q-1}\).

\(\frac{P^2+3}{Q-1}\). The condition \(Q\ne1\) guarantees a nonzero denominator limit, which is required for the quotient law.

- Find the limit of the expression inside the radical first. - Check that the outer function behaves continuously at that limiting input. - Then pass the limiting value through the outer function.

1. By the sum law, \(\lim_{x\to5}(h(x)+7)=9+7=16\). 2. The square-root function is continuous at \(16\). 3. Therefore \(\lim_{x\to5}\sqrt{h(x)+7}=\sqrt{16}=4\).

\(4\). The inner expression approaches \(16\), where the square-root function is continuous.

- Evaluate each polynomial factor at the approach value. - Keep the two factor limits separate until both are simplified. - Apply the product law after confirming both finite limits exist.

1. Polynomials are continuous, so direct substitution gives \(\lim_{x\to2}(x^2-3x+4)=2^2-3\cdot2+4=2\). 2. Likewise, \(\lim_{x\to2}(2x-1)=2\cdot2-1=3\). 3. By the product law, the required limit is \(2\cdot3=6\).

\(6\).

- Find the limit of the quantity inside the radical first. - Recall the real domain of the cube-root function. - Apply the outer function to the inner limiting value once continuity is established.

1. By the sum law, \(\lim_{x\to2}(f(x)+g(x))=-3+(-5)=-8\). 2. The cube-root function is continuous for every real input, including negative inputs. 3. Therefore \(\lim_{x\to2}\sqrt[3]{f(x)+g(x)}=\sqrt[3]{-8}=-2\).

\(-2\). The cube-root function is continuous on all real numbers, so a negative inner limit causes no difficulty.

- Find the numerator and denominator limits independently. - Check the denominator limit before forming the quotient. - Keep track of the signs when simplifying the final ratio.

1. The numerator limit is \(1\cdot 2+(-4)=-2\) by the product and sum laws. 2. The denominator limit is \(1+(-4)=-3\), which is nonzero. 3. The quotient law gives \(\frac{-2}{-3}=\frac{2}{3}\).

\(\frac{2}{3}\).

- Substitute the approach value into the numerator and denominator separately. - Check the denominator value before using the quotient law. - Form the quotient only after confirming the denominator limit is nonzero.

1. The numerator is a polynomial, so its limit is \((-1)^2+4(-1)+7=4\). 2. The denominator limit is \(2(-1)^2+3(-1)+4=3\). 3. Because the denominator limit is nonzero, the quotient law applies and the limit is \(\frac{4}{3}\).

\(\frac{4}{3}\).

- Think of the whole expression as built from sums, constant multiples, and a power of \(f(x)\). - Apply the known limiting value to each component. - Combine the component limits only after evaluating them separately.

1. The power law gives \(\lim_{x\to3}[f(x)]^3=(-2)^3=-8\). 2. The constant multiple law gives \(\lim_{x\to3}[-4f(x)]=-4\cdot(-2)=8\). 3. Combining the terms gives \(-8+8+1=1\).

\(1\).

- Apply the limit laws to the numerator exactly as written. - Evaluate the denominator limit and check that it is nonzero. - The expression can be handled law by law without factoring.

1. The numerator limit is \(5^2-2^2=25-4=21\) by the power and difference laws. 2. The denominator limit is \(5+2=7\), which is nonzero. 3. By the quotient law, the limit is \(\frac{21}{7}=3\).

\(3\).

- Treat the two parenthesized expressions as separate factors. - Use the sum and difference laws inside each factor. - Apply the product law only after both factor limits are known.

1. The first factor approaches \(2-(-1)=3\). 2. The second factor approaches \(-1+3=2\). 3. By the product law, the required limit is \(3\cdot 2=6\).

\(6\).

- Handle the inner quotient before the outer square. - Check the denominator limit before using the quotient law. - Apply the power law only after the quotient's limit is known.

1. Since \(\lim_{x\to3}g(x)=\frac{1}{2}\ne0\), the quotient law gives \(\lim_{x\to3}\frac{f(x)}{g(x)}=\frac{-2}{\frac{1}{2}}=-4\). 2. Apply the power law to the quotient limit. 3. The required limit is \((-4)^2=16\).

\(16\).

- Match the operation between the functions to the corresponding limit law. - Keep the structure of the original expression unchanged when combining limits. - Compare multiplication with addition before calculating.

1. The expression is a product, so the product law applies rather than the sum law. 2. The product law gives \(\lim_{x\to4}f(x)g(x)=2\cdot 5\). 3. Therefore the correct limit is \(10\), not \(7\).

The student used the sum law on a product. The correct limit is \(10\).

- Apply the limit laws separately to the numerator and denominator. - Evaluate the power before adding the constant in the denominator limit. - Check that the denominator's limiting value allows the quotient law.

1. The numerator approaches \(2\cdot(-1)+3=1\). 2. The denominator approaches \(2^2+1=5\), which is nonzero. 3. By the quotient law, the limit is \(\frac{1}{5}\).

\(\frac{1}{5}\).

- Check the common denominator's limiting value once. - Evaluate the two quotient limits separately. - Add their limiting values at the end.

1. Since \(h(x)\to2\ne0\), both quotient limits are valid. 2. The first quotient approaches \(\frac{3}{2}\), and the second approaches \(\frac{4}{2}=2\). 3. By the sum law, the total limit is \(\frac{3}{2}+2=\frac{7}{2}\).

\(\frac{7}{2}\).

- Use continuity to substitute the angle into sine and cosine. - Recall the exact values at \(\frac{\pi}{3}\). - Apply the constant-multiple and difference laws before simplifying.

1. The sine and cosine functions are continuous at \(\frac{\pi}{3}\). 2. Direct substitution gives \(2\sin\left(\frac{\pi}{3}\right)-\cos\left(\frac{\pi}{3}\right)=2\cdot\frac{\sqrt{3}}{2}-\frac{1}{2}\). 3. Therefore the limit is \(\sqrt{3}-\frac{1}{2}\).

\(\sqrt{3}-\frac{1}{2}\).

- Match the operation in the original expression to the correct limit law. - Keep the exponent attached to the function's limiting value. - Pay attention to the effect of an even power on a negative number.

1. The power law says the limiting value is raised to the fourth power; the exponent is not moved in front. 2. Thus \(\lim_{x\to8}[r(x)]^4=(-2)^4\). 3. The correct limit is \(16\).

The student confused a power with multiplication by the exponent. The correct limit is \(16\).

- Evaluate each form independently using the operations it displays. - For the factored form, find both factor limits before multiplying. - Algebraically equivalent expressions should agree wherever the relevant limit laws apply.

1. The first form has limit \((-1)^2-3^2=1-9=-8\). 2. For the factored form, the two factors approach \(-1-3=-4\) and \(-1+3=2\). 3. Their product limit is \((-4)\cdot 2=-8\), matching the first form.

Both limits equal \(-8\).

- Apply the law matching each operation separately. - Zero causes no problem for sums or products. - Check the extra condition required for quotients.

1. By the sum law, part a has limit \(0+0=0\). 2. By the product law, part b has limit \(0\cdot 0=0\). 3. For part c, the denominator approaches \(0\), so the quotient law does not apply. The quotient limit is not determined by the two given limits alone.

a) \(0\) b) \(0\) c) Not determined from the given information.

- The quotient law requires a nonzero denominator limit. - Keep the numerator simple and vary how the denominator approaches \(0\). - Compare a denominator that stays positive with one that changes sign.

1. The quotient law cannot be applied because the denominator limit is \(0\). The symbol \(\frac{-1}{0}\) is not a numerical limit calculation. 2. Let \(f(x)=-1\) and \(g_1(x)=(x-2)^2\). Then the quotient is \(-\frac{1}{(x-2)^2}\), which tends to \(-\infty\) from both sides. 3. Let \(f(x)=-1\) and \(g_2(x)=x-2\). Then the quotient is \(-\frac{1}{x-2}\), which tends to \(+\infty\) from the left and \(-\infty\) from the right, so its two-sided limit does not exist. 4. Both denominator functions tend to \(0\), but the quotient behavior differs. Therefore the original limit information alone is insufficient.

The conclusion is not justified. The denominator approaching \(0\) does not determine the quotient limit. For example, \(g(x)=(x-2)^2\) gives a quotient tending to \(-\infty\), while \(g(x)=x-2\) gives unequal one-sided infinite behavior and no two-sided limit.

- Translate the limit of the sum into a relationship among the individual limits. - Represent the unknown limit with a symbol. - Use the known limit of \(g\) to isolate the remaining one.

1. By the sum and constant multiple laws, \(\lim_{x\to6}(f(x)+2g(x))=\lim_{x\to6}f(x)+2\lim_{x\to6}g(x)\). 2. Let \(L=\lim_{x\to6}f(x)\). Then \(L+2\cdot 4=11\). 3. Solving gives \(L=3\).

\(\lim_{x\to6}f(x)=3\).

- Apply the laws separately to each expression rather than assuming the same rule works for all three. - For a quotient, inspect the denominator's limiting value before dividing. - A zero denominator limit signals that additional local information may be needed.

1. For a), the constant multiple and difference laws give \(3\cdot 2-(-3)=9\). 2. For b), the power and sum laws give \(2^2+(-3)=1\). 3. For c), the denominator approaches \(2-2=0\). The quotient law therefore does not apply, and the given information does not determine the quotient limit.

a) \(9\) b) \(1\) c) Not determined from the given information because the denominator approaches \(0\).

- Replace the function limits with their known values while keeping \(k\) symbolic. - Check the denominator limit before using the quotient law. - Set the resulting numerical expression equal to the required limit value.

1. The numerator limit is \(3k+1\), and the denominator limit is \(3-1=2\). 2. Since the denominator limit is nonzero, the quotient law gives \(\frac{3k+1}{2}=2\). 3. Thus \(3k+1=4\), so \(k=1\).

\(k=1\).

- Replace each occurrence of \(f(x)\) by the same unknown limiting value. - Use the stated condition to justify the quotient law. - Solve the resulting algebraic equation for the unknown limit.

1. Since \(L\ne1\), the quotient law gives \(\frac{L+1}{L-1}=3\). 2. Multiply by \(L-1\): \(L+1=3L-3\). 3. Thus \(4=2L\), so \(L=2\).

\(\lim_{x\to5}f(x)=2\).

- Represent the two unknown limits with two variables. - Translate each given combined limit into an equation. - Solve the resulting pair of equations together.

1. Let \(F=\lim_{x\to2}f(x)\) and \(G=\lim_{x\to2}g(x)\). 2. The sum and difference laws give \(F+G=7\) and \(F-G=1\). 3. Solving the system gives \(F=4\) and \(G=3\).

\(\lim_{x\to2}f(x)=4\) and \(\lim_{x\to2}g(x)=3\).

- Rewrite the unknown function using the quotient whose limit is known. - Treat that quotient as one factor and \(g(x)\) as the other. - Apply the product law to the two known factor limits.

1. Near the target, \(f(x)=\left(\frac{f(x)}{g(x)}\right)g(x)\). 2. The two factors have limits \(4\) and \(3\). 3. By the product law, \(\lim_{x\to-2}f(x)=4\cdot 3=12\).

\(\lim_{x\to-2}f(x)=12\).