Free math worksheets

- Determine the nearby limiting value from the rule used away from the point. - Compare that limit with the assigned point value. - Ask whether changing only one function value could restore continuity.

1. For nearby inputs, \(g(x)=x^2\), so \(\lim_{x\to-2}g(x)=4\). 2. The actual value is \(g(-2)=1\), which does not equal the limit. 3. The discontinuity is removable. Redefining \(g(-2)=4\) would make the function continuous at \(-2\).

The discontinuity is removable, and the repairing value is \(g(-2)=4\).

- “Started” means round the number of \(100\,\text{cm}^3\) blocks up. - Determine whether the fee changes continuously or in fixed jumps. - Identify the multiples of \(100\) in the interval.

1. A displacement of \(1200\,\text{cm}^3\) uses exactly \(12\) blocks, so \(H(1200)=12\cdot\$2.00=\$24.00\). 2. Both \(1201\,\text{cm}^3\) and \(1299\,\text{cm}^3\) start a thirteenth block, so each fee is \(13\cdot\$2.00=\$26.00\). 3. The graph is a step function. It stays constant within each \(100\,\text{cm}^3\) block and jumps immediately after each multiple of \(100\). 4. Therefore, the function is discontinuous at \(V=1100, 1200, 1300, 1400\) within the stated interval.

a) \(H(1200)=\$24.00\), \(H(1201)=\$26.00\), and \(H(1299)=\$26.00\) b) \(H\) is a step function and is not continuous. The jump discontinuities are at \(V=1100, 1200, 1300, 1400\).

- What result do you get from direct substitution? - Factor the numerator or denominator to look for a common factor. - If the factor causing the discontinuity cancels completely, what type of discontinuity remains? - If an uncanceled denominator factor approaches zero, compare the behavior from the left and right.

1. For part a, direct substitution gives \(\frac{0}{0}\). Factor: \(f(x)=\frac{(x-3)(x+3)}{2(x-3)}\). 2. For \(x\neq 3\), \(f(x)=\frac{x+3}{2}\). Therefore, \(\lim_{x\to 3}f(x)=\frac{3+3}{2}=3\). The finite limit shows that \(x=3\) is a removable discontinuity. 3. For part b, direct substitution also gives \(\frac{0}{0}\). Factor: \(g(x)=\frac{x-1}{(x-1)^2}\). 4. For \(x\neq 1\), \(g(x)=\frac{1}{x-1}\). As \(x\to 1^-\), \(g(x)\to -\infty\), and as \(x\to 1^+\), \(g(x)\to +\infty\). 5. The two-sided limit does not exist, and \(x=1\) is an infinite discontinuity.

a) \(\lim_{x\to 3}f(x)=3\); the discontinuity is removable. b) \(\lim_{x\to 1}g(x)\) does not exist because the one-sided limits are \(-\infty\) and \(+\infty\); the discontinuity is infinite.

- Evaluate the left-hand and right-hand limits using different branches. - Compare the two one-sided limiting values before looking at the point value. - A finite mismatch between the sides has a specific discontinuity classification.

1. From the left, \(f(x)=x+2\), so \(\lim_{x\to1^-}f(x)=3\). 2. From the right, \(f(x)=5-x\), so \(\lim_{x\to1^+}f(x)=4\). 3. The one-sided limits are finite but unequal, so the two-sided limit does not exist. The discontinuity at \(x=1\) is a jump discontinuity; the value \(f(1)=7\) does not change that classification.

\(f\) is not continuous at \(x=1\). It has a jump discontinuity there.

- Evaluate each branch at the boundary value through a one-sided limit. - Check whether the two sides approach the same number. - If only the point value is missing, consider whether one assignment can repair continuity.

1. The left-hand limit is \(3\cdot 2-1=5\). 2. The right-hand limit is \(2^2+1=5\). 3. The two-sided limit exists and equals \(5\), but the function is undefined at \(2\). The discontinuity is removable, and defining \(h(2)=5\) would remove it.

Removable discontinuity; define \(h(2)=5\).

- Examine the sign of the squared denominator on both sides of \(2\). - Determine what happens when a positive denominator approaches zero. - Match unbounded nearby behavior to its discontinuity type.

1. On either side of \(2\), \((x-2)^2\) is positive and approaches \(0\). 2. Therefore \(p(x)\to+\infty\) from both sides. 3. The function has an infinite discontinuity at \(x=2\), with vertical asymptote \(x=2\).

\(p(x)\to+\infty\) from both sides. The discontinuity is infinite, and \(x=2\) is a vertical asymptote.

- Determine what the squared logarithm input approaches on each side of \(-1\). - Recall the behavior of \(\ln u\) as a positive input \(u\) approaches \(0\). - Unbounded behavior at a finite input determines the discontinuity type.

1. On both sides of \(x=-1\), the quantity \((x+1)^2\) is positive and approaches \(0\). 2. Because \(\ln u\to-\infty\) as \(u\to0^+\), \(q(x)\to-\infty\) from the left. 3. The same argument gives \(q(x)\to-\infty\) from the right. 4. The discontinuity is infinite, and \(x=-1\) is a vertical asymptote.

\(\lim_{x\to-1^-}q(x)=-\infty\) and \(\lim_{x\to-1^+}q(x)=-\infty\). The discontinuity is infinite.

- Separate “undefined at the point” from “unbounded near the point.” - Determine the nearby limit before naming the discontinuity type. - Ask whether assigning one finite point value would repair continuity.

1. The function is undefined at \(x=0\), but an excluded input does not by itself imply unbounded behavior. 2. The standard limit \(\lim_{x\to0}\frac{\sin x}{x}=1\) is finite. 3. Therefore the discontinuity at \(x=0\) is removable, not infinite. Defining \(g(0)=1\) would make the function continuous there.

The claim is false. The discontinuity is removable, and the continuous extension uses \(g(0)=1\).

- Separate the table into inputs less than and greater than \(4\). - Track both the sign and magnitude of the outputs on each side. - Match unbounded nearby behavior to the appropriate discontinuity type.

1. From the left of \(4\), the outputs become negative with rapidly increasing magnitude, suggesting \(h(x)\to-\infty\). 2. From the right, the outputs become positive with rapidly increasing magnitude, suggesting \(h(x)\to+\infty\). 3. The unbounded one-sided behavior indicates an infinite discontinuity at \(x=4\).

Infinite discontinuity at \(x=4\); the table suggests \(h(x)\to-\infty\) from the left and \(h(x)\to+\infty\) from the right.

- Rewrite tangent as a quotient of sine and cosine. - Track the sign of cosine on each side of \(\frac{\pi}{2}\). - Unbounded one-sided behavior identifies the discontinuity type.

1. As \(x\to\frac{\pi}{2}^-\), \(\cos x\to0^+\) while \(\sin x\to1\), so \(\tan x\to+\infty\). 2. As \(x\to\frac{\pi}{2}^+\), \(\cos x\to0^-\) while \(\sin x\to1\), so \(\tan x\to-\infty\). 3. The function has an infinite discontinuity at \(x=\frac{\pi}{2}\), which is a vertical asymptote.

Infinite discontinuity at \(x=\frac{\pi}{2}\); the left-hand limit is \(+\infty\) and the right-hand limit is \(-\infty\).

- Rewrite \(|x|\) separately for negative and positive inputs. - Find each one-sided constant value. - Classify the discontinuity from the relationship between the two finite one-sided limits.

1. For \(x<0\), \(|x|=-x\), so \(s(x)=-1\). Thus \(\lim_{x\to0^-}s(x)=-1\). 2. For \(x>0\), \(|x|=x\), so \(s(x)=1\). Thus \(\lim_{x\to0^+}s(x)=1\). 3. The one-sided limits are finite but unequal, so \(x=0\) is a jump discontinuity.

\(\lim_{x\to0^-}s(x)=-1\), \(\lim_{x\to0^+}s(x)=1\); jump discontinuity at \(x=0\).

- Analyze nearby behavior before considering the assigned point value. - Ask whether the two-sided limit is finite. - A removable discontinuity can be repaired at one point; unbounded behavior cannot.

1. As \(x\to1\) from either side, \((x-1)^2\to0^+\), so \(\frac{1}{(x-1)^2}\to+\infty\). 2. The nearby behavior is unbounded, so no finite choice of \(r(1)\) can make the function continuous at \(1\). 3. The discontinuity is infinite, not removable.

The claim is false. The discontinuity at \(x=1\) is infinite, and assigning a finite point value cannot remove it.

- Estimate each one-sided limit separately from the table. - Continuity requires agreement from both sides, not only one side and the point value. - Classify the discontinuity from the relationship between the one-sided limits.

1. The left-side values approach \(6\), matching \(v(-2)=6\). 2. The right-side values approach \(1\). 3. Since the finite one-sided limits are unequal, the two-sided limit does not exist. The discontinuity is a jump discontinuity regardless of the point value.

Jump discontinuity at \(x=-2\). The right-hand limit approaches \(1\), so matching the left-hand limit and point value is not enough for continuity.

- Compare the left-hand and right-hand behavior in each panel before considering any filled point. - A finite common limit with a mismatched point value is removable; unequal finite one-sided limits form a jump. - Unbounded behavior on either side indicates an infinite discontinuity.

1. In panel a), both sides approach \(1\), while the filled point is at \((0,3)\). The discontinuity is removable. 2. In panel b), the left side approaches \(1\) and the right side approaches \(3\). The discontinuity is a jump. 3. In panel c), both branches increase without bound as \(x\to0\). The discontinuity is infinite.

a) Removable b) Jump c) Infinite

- Use the domain to identify the available direction of approach. - Focus on the logarithm's input \(x-2\). - Recall the logarithm's behavior as its input approaches \(0\) from the positive side.

1. As \(x\to2^+\), \(x-2\to0^+\). 2. The natural logarithm decreases without bound as its positive input approaches \(0\), so \(w(x)\to-\infty\). 3. Thus \(x=2\) is a vertical asymptote, corresponding to infinite discontinuity behavior at the domain boundary.

\(w(x)\to-\infty\) as \(x\to2^+\). The graph has vertical asymptote \(x=2\), with infinite discontinuity behavior there.

- Decide whether a two-sided limit exists before choosing a point value. - Redefining one point changes only the function value, not nearby one-sided behavior. - A repair by point redefinition is possible only for one discontinuity type.

1. The left-hand and right-hand limits are finite but unequal. 2. Therefore the two-sided limit at \(a\) does not exist. 3. Changing only \(g(a)\) cannot change either one-sided limit, so no choice of \(c\) makes the function continuous. The discontinuity is a jump.

No real value of \(c\) can make \(g\) continuous at \(a\). The discontinuity is a jump.

- Simplify the nearby formula while preserving the separate point definition. - Compare the nearby limit with the assigned value at \(2\). - A function can be defined specifically to fill a hole that another formula would have had.

1. For \(x\ne2\), \(\frac{x^2-4}{x-2}=x+2\). 2. Therefore \(\lim_{x\to2}F(x)=4\). 3. Since the defined value is also \(F(2)=4\), the continuity condition is satisfied. The removable discontinuity has already been repaired, so no discontinuity remains at \(x=2\).

No discontinuity remains at \(x=2\); \(F\) is continuous there.

- Check whether the original function has a value at \(x=1\). - Distinguish a limit from an actual function value. - Canceling a factor does not restore an excluded input automatically.

1. The original denominator is \(0\) at \(x=1\), so \(f(1)\) is not defined. 2. Factoring gives \(\frac{(x-1)(x+1)}{x-1}=x+1\) only for \(x\ne1\). 3. Although \(\lim_{x\to1}f(x)=2\), continuity at \(x=1\) requires the function to be defined there and to satisfy \(f(1)=2\). 4. Therefore, the statement is false. The graph has a removable discontinuity, or hole, at \((1, 2)\).

The statement is false. The limit is \(2\), but \(f(1)\) is undefined, so \(f\) is not continuous at \(x=1\). The discontinuity is removable.

- Find the input that makes the denominator \(0\). - Factor and simplify for allowed inputs. - Use the simplified rule to find the missing limiting value.

1. The denominator is \(0\) at \(x=3\), so the domain is \(\mathbb{R}\setminus\{3\}\). 2. Factor the numerator: \(x^2-4x+3=(x-3)(x-1)\). For \(x\ne3\), \(g(x)=x-1\). 3. Therefore, \(\lim_{x\to3}g(x)=\lim_{x\to3}(x-1)=2\). The function values approach \(2\) from both sides. 4. The discontinuity is removable. Defining \(g(3)=2\) makes the extended function continuous.

a) \(\mathbb{R}\setminus\{3\}\) b) The function values approach \(2\) as \(x\to3\). c) Yes. Define \(g(3)=2\).

- Pay attention to whether each boundary value is included in the lower or upper interval. - Use piecewise notation to show a different constant value on each weight interval. - Continuity requires the two one-sided limits to be equal.

1. The package weighing \(1.8\,\text{lb}\) is in the first interval, so its cost is \(\$4.95\). The package weighing \(5.0\,\text{lb}\) is included in the second interval, so its cost is \(\$6.95\). The package weighing \(5.1\,\text{lb}\) is in the third interval, so its cost is \(\$10.45\). 2. The cost function is \(K(w)=\begin{cases}\$4.95&\text{if }0<w\le2\\\$6.95&\text{if }2<w\le5\\\$10.45&\text{if }5<w\le10\\\$18.45&\text{if }10<w\le31.5\end{cases}\). 3. At \(w=5\), \(\lim_{w\to5^-}K(w)=\$6.95\), while \(\lim_{w\to5^+}K(w)=\$10.45\). 4. Since the one-sided limits are different, the two-sided limit does not exist. The function has a jump discontinuity at \(w=5\).

a) \(1.8\,\text{lb}:\ \$4.95\); \(5.0\,\text{lb}:\ \$6.95\); \(5.1\,\text{lb}:\ \$10.45\) b) \(K(w)=\begin{cases}\$4.95&0<w\le2\\\$6.95&2<w\le5\\\$10.45&5<w\le10\\\$18.45&10<w\le31.5\end{cases}\) c) Not continuous, because \(\lim_{w\to5^-}K(w)=\$6.95\ne\$10.45=\lim_{w\to5^+}K(w)\).

- Find the zeros of the denominator to determine the excluded values. - Factor the numerator and denominator before evaluating the limits. - What does a canceled factor imply about the corresponding discontinuity? - Compare the left-hand and right-hand behavior near an uncanceled denominator zero.

1. Factor the denominator: \(x^2-3x+2=(x-1)(x-2)\). Therefore, \(D_f=\mathbb{R}\setminus\{1, 2\}\). 2. Factor the numerator: \(x^2-1=(x-1)(x+1)\). For \(x\neq 1\), \(f(x)=\frac{x+1}{x-2}\). 3. Evaluate the first limit: \(\lim_{x\to 1}f(x)=\frac{1+1}{1-2}=-2\). Because the limit is finite, \(x=1\) is a removable discontinuity. 4. Near \(x=2\), the numerator of the simplified expression approaches \(3\), while the denominator approaches \(0\). Specifically, \(\lim_{x\to 2^-}f(x)=-\infty\) and \(\lim_{x\to 2^+}f(x)=+\infty\). 5. The one-sided limits are different, so \(\lim_{x\to 2}f(x)\) does not exist. The discontinuity at \(x=2\) is infinite.

a) \(D_f=\mathbb{R}\setminus\{1, 2\}\) b) \(\lim_{x\to 1}f(x)=-2\). The limit \(\lim_{x\to 2}f(x)\) does not exist because the one-sided limits are \(-\infty\) and \(+\infty\). c) The discontinuity at \(x=1\) is removable, and the discontinuity at \(x=2\) is infinite.

- Try direct substitution first. - Factor the numerator and denominator when you obtain \(\frac{0}{0}\). - Use one-sided signs when a denominator approaches zero without a canceling factor.

1. Direct substitution gives \(f(0)=\frac{-8}{-16}=\frac{1}{2}\). This is the y-intercept. 2. Factor: \(f(x)=\frac{(x-4)(x+2)}{(x-4)(x+4)}\). For \(x\ne4\), this simplifies to \(\frac{x+2}{x+4}\). Therefore, the limit at \(4\) is \(\frac{6}{8}=\frac{3}{4}\). The original graph has a removable discontinuity, or hole, at \(\left(4, \frac{3}{4}\right)\). 3. Near \(x=-4\), the simplified numerator approaches \(-2\), while the denominator approaches \(0\). From the left, the quotient approaches \(+\infty\); from the right, it approaches \(-\infty\). Therefore, the two-sided limit does not exist, and \(x=-4\) is a vertical asymptote.

1. \(\frac{1}{2}\) 2. \(\frac{3}{4}\); removable discontinuity at \(\left(4, \frac{3}{4}\right)\) 3. The two-sided limit does not exist: the left-hand limit is \(+\infty\), and the right-hand limit is \(-\infty\). The vertical asymptote is \(x=-4\).

- Factor the denominator to find all excluded values. - Factor the numerator and check for common factors. - If the denominator has greater multiplicity than the numerator at a common zero, what remains after simplification? - Use the simplified expression to evaluate the behavior near each excluded value.

1. Factor the denominator: \(x^3-2x^2=x^2(x-2)\). The discontinuities occur at \(x=0\) and \(x=2\). 2. Factor the numerator: \(x^2-4=(x-2)(x+2)\). 3. At \(x=2\), cancel the common factor for \(x\neq 2\): \(g(x)=\frac{x+2}{x^2}\). Then \(\lim_{x\to 2}g(x)=\frac{2+2}{2^2}=1\). Thus, \(x=2\) is a removable discontinuity. 4. At \(x=0\), use the simplified expression \(\frac{x+2}{x^2}\). The numerator approaches \(2\), and \(x^2\to 0^+\) from both sides. Therefore, \(\lim_{x\to 0}g(x)=+\infty\). 5. Thus, \(x=0\) is an infinite discontinuity, the graph does not change sign across it, and the vertical asymptote is \(x=0\).

a) The discontinuities are \(x=0\) and \(x=2\). b) \(\lim_{x\to 2}g(x)=1\), and \(\lim_{x\to 0}g(x)=+\infty\). c) The discontinuity at \(x=2\) is removable. The discontinuity at \(x=0\) is infinite, with vertical asymptote \(x=0\).

- A filled point gives the actual function value; an open circle is excluded. - Look for places where the graph has a hole or a jump. - Compare the values approached from the left and right at each break. - For a removable discontinuity, the limit exists but differs from the function value.

1. At \(x=-2\), the filled point is at \((-2, 3)\), so \(f(-2)=3\). The open circle at \((-2, 1)\) is not included. 2. At \(x=1\), the filled point is at \((1, 0)\), so \(f(1)=0\). The open circle at \((1, 2.5)\) is not included. 3. At \(x=3\), the graph lies on the line segment at \(y=-2\), so \(f(3)=-2\). 4. At \(x=-2\), both one-sided limits equal \(1\), but \(f(-2)=3\). Therefore, \(f\) has a removable discontinuity there. 5. At \(x=1\), the left-hand limit is \(2.5\), while the right-hand limit is \(0\). Therefore, \(f\) has a jump discontinuity there. The graph is continuous at all other inputs in the interval.

a) \(f(-2)=3\), \(f(1)=0\), and \(f(3)=-2\) b) \(f\) is not continuous at \(x=-2\) and \(x=1\). At \(x=-2\), the discontinuity is removable; at \(x=1\), it is a jump discontinuity.

- First determine how the horizontal shift changes the input of the discontinuity. - Reflection across the x-axis changes output values but not input values. - Read the original one-sided limits and function value at \(x=1\). - Multiply the relevant output values by \(-1\).

1. Shifting the graph \(2\) units to the right moves the discontinuity from \(x=1\) to \(x_0=3\). Reflecting across the x-axis does not change its x-coordinate. 2. For the original function, \(\lim_{x\to1^-}f(x)=0\), \(\lim_{x\to1^+}f(x)=1\), and \(f(1)=1\). 3. Reflection across the x-axis multiplies every output by \(-1\). Therefore, \(\lim_{x\to3^-}h(x)=0\), \(\lim_{x\to3^+}h(x)=-1\), and \(h(3)=-1\).

a) \(x_0=3\) b) \(\lim_{x\to3^-}h(x)=0\), \(\lim_{x\to3^+}h(x)=-1\), and \(h(3)=-1\)

- Use filled points to identify actual function values. - Use open circles and nearby graph pieces to identify limiting values. - Compare the left-hand limit, right-hand limit, and function value at each input.

1. At \(x=-1\), the graph approaches \(1\) from both sides, so \(\lim_{x\to-1}g(x)=1\). However, the filled point shows that \(g(-1)=2\). Since the limit and function value are different, \(g\) has a removable discontinuity at \(x=-1\). 2. At \(x=2\), the left-hand limit is \(4\), the right-hand limit is \(0\), and the filled point gives \(g(2)=4\). Since the one-sided limits are different, \(g\) has a jump discontinuity at \(x=2\).

At \(x=-1\), \(g\) is not continuous; it has a removable discontinuity. At \(x=2\), \(g\) is not continuous; it has a jump discontinuity.

- Reflection across the y-axis changes the sign of each x-coordinate. - Locate the jump in the original graph first. - Reflection across the y-axis reverses the left and right sides of approach. - Compare the two one-sided limits at the reflected input.

1. The graph of \(f\) has a jump at \(x_f=1\): the left-hand limit is \(0\), while the right-hand limit is \(2\). 2. Reflection across the y-axis changes each input \(x\) to \(-x\), so the jump moves from \(x=1\) to \(x_g=-1\). 3. The reflection reverses the sides of approach. Thus \(\lim_{x\to-1^-}g(x)=2\) and \(\lim_{x\to-1^+}g(x)=0\). 4. The filled point on the original graph gives \(f(1)=0\), so \(g(-1)=f(1)=0\). Since the one-sided limits are different, \(g\) is not continuous at \(x=-1\).

a) \(x_f=1\) b) \(x_g=-1\) c) \(\lim_{x\to-1^-}g(x)=2\) and \(\lim_{x\to-1^+}g(x)=0\). Since the one-sided limits are different, \(g\) has a jump discontinuity at \(x=-1\).

- Track how the horizontal translation moves the original jump. - Use the filled point for the function value. - Read the left and right branches separately for the one-sided limits. - Apply the vertical shift to all three y-values.

1. Replacing \(x\) with \(x-3\) shifts the graph right \(3\) units, and adding \(1\) shifts it up \(1\) unit. 2. The jump at \(x=1\) therefore moves to \(x_0=1+3=4\). 3. The filled point on the original graph is \((1, -5)\), so \(k(4)=f(1)+1=-5+1=-4\). 4. The left-hand limit of \(f\) at \(1\) is \(-5\), so \(\lim_{x\to4^-}k(x)=-5+1=-4\). 5. The right-hand limit of \(f\) at \(1\) is \(-1\), so \(\lim_{x\to4^+}k(x)=-1+1=0\).

a) Shift right \(3\) units and up \(1\) unit. b) \(x_0=4\) c) \(k(4)=-4\), \(\lim_{x\to4^-}k(x)=-4\), and \(\lim_{x\to4^+}k(x)=0\)

- An open circle shows a value the graph approaches but does not include. - A filled point gives the actual function value. - Continuity requires the limit to equal the function value.

1. As \(x\) approaches \(2\) from either side, the graph approaches \(3\). Therefore, \(\lim_{x\to2}g(x)=3\). 2. The filled point gives the actual function value \(g(2)=1\). 3. Since \(\lim_{x\to2}g(x)\ne g(2)\), the function has a removable discontinuity at \(x=2\). 4. To make the function continuous, the missing point on the line must be filled, so the function must be redefined with \(g(2)=3\).

a) \(g\) is not continuous at \(x=2\) because \(\lim_{x\to2}g(x)=3\), but \(g(2)=1\). b) \(g(2)=3\)

- Analyze each boundary independently. - At each boundary, compare the left-hand limit, right-hand limit, and point value. - A matching limit with the wrong point value differs fundamentally from unequal one-sided limits.

1. At \(x=0\), the left-hand limit is \(1\) and the right-hand limit is also \(1\), but \(f(0)=5\). Thus \(x=0\) is a removable discontinuity. 2. At \(x=2\), the left-hand limit from \(1-x\) is \(-1\). 3. From the right, the branch \(x\) approaches \(2\), so the one-sided limits differ. Thus \(x=2\) is a jump discontinuity.

At \(x=0\): removable discontinuity. At \(x=2\): jump discontinuity.

- First solve for the x-value that sends the transformed input to the original discontinuity. - Transform the original limiting value and point value separately. - Decide whether the transformed mismatch can still be fixed by changing one point value.

1. The input of \(f\) equals \(-1\) when \(x+3=-1\), so the transformed discontinuity occurs at \(x=-4\). 2. As \(x\to-4\), \(f(x+3)\to4\), so \(g(x)\to2\cdot 4-5=3\). The actual value is \(g(-4)=2\cdot 9-5=13\). 3. The discontinuity remains removable. Redefining \(g(-4)=3\) would make \(g\) continuous there.

The discontinuity is removable at \(x=-4\), and the repairing value is \(g(-4)=3\).

- Determine the behavior of the exponent \(1/x\) separately from each side. - Translate those exponent limits through the exponential function. - A discontinuity can be classified as infinite even when only one side is unbounded.

1. As \(x\to0^-\), \(1/x\to-\infty\), so \(e^{1/x}\to0\). 2. As \(x\to0^+\), \(1/x\to+\infty\), so \(e^{1/x}\to+\infty\). 3. Because the function is unbounded on one side of \(0\), it has an infinite discontinuity there.

\(\lim_{x\to0^-}u(x)=0\), while \(u(x)\to+\infty\) as \(x\to0^+\). The discontinuity is infinite.

- Find each one-sided limit in terms of \(k\). - Separate the case where the one-sided limits agree from the case where they do not. - Only after the sides agree should you compare the common limit with the point value.

1. The left-hand limit is always \(0\). 2. The right-hand limit is \(k\). 3. If \(k=0\), the two-sided limit is \(0\) but \(f_k(0)=2\), so the discontinuity is removable. If \(k\ne0\), the finite one-sided limits differ, so the discontinuity is a jump.

If \(k=0\): removable discontinuity. If \(k\ne0\): jump discontinuity.

- Analyze the two target inputs separately. - At \(0\), focus on unbounded behavior of the reciprocal branch. - At \(1\), compare the finite boundary values approached by the two branches.

1. Near \(x=0\), the nearby rule is \(1/x\). It approaches \(-\infty\) from the left and \(+\infty\) from the right, so \(x=0\) is an infinite discontinuity. The assigned value \(R(0)=0\) does not change this. 2. At \(x=1\), the left-hand limit from \(1/x\) is \(1\). 3. The right-hand limit from \(x+2\) is \(3\). Since these finite limits differ, \(x=1\) is a jump discontinuity.

At \(x=0\): infinite discontinuity. At \(x=1\): jump discontinuity.

- Factor both the numerator and denominator before classifying any singularity. - Check separately whether the second denominator factor also vanishes at \(x=2\). - After setting \(a=1\), examine every zero of the original denominator.

1. Factor the expression: \(x^2-4=(x-2)(x+2)\) and \(x^2+(a-2)x-2a=(x-2)(x+a)\). 2. If \(a\ne-2\), canceling \(x-2\) gives \(\frac{x+2}{x+a}\) near \(x=2\), so the finite limit is \(\frac{4}{a+2}\). Thus the discontinuity at \(x=2\) is removable, and defining \(F_a(2)=\frac{4}{a+2}\) makes the extension continuous there. 3. If \(a=-2\), the denominator is \((x-2)^2\), so the simplified nearby expression is \(\frac{x+2}{x-2}\). It is unbounded at \(x=2\), making the discontinuity infinite and \(x=2\) a vertical asymptote. 4. For \(a=1\), the original denominator is \((x-2)(x+1)\). The point \(x=2\) is removable with limit \(\frac{4}{3}\), while \(x=-1\) is an infinite discontinuity and the vertical asymptote is \(x=-1\).

a) If \(a\ne-2\), \(x=2\) is removable. If \(a=-2\), it is infinite. b) Define \(F_a(2)=\frac{4}{a+2}\) for \(a\ne-2\). c) For \(a=1\), \(x=2\) is removable, \(x=-1\) is infinite, and the vertical asymptote is \(x=-1\).