Removing Discontinuities - Calculus 1

1. Let $f(x)$ be a function with a removable discontinuity at $x=c$, and let $g(x)$ be a function that is continuous at $x=c$. Let $h(x) = f(x)g(x)$. Which condition on $g(x)$ is sufficient to guarantee that $h(x)$ is continuous at $x=c$?
   Answer: $g(c) = 0$ When analyzing products of functions with discontinuities, you need to consider how the behavior of each function affects the overall continuity. Since $f(x)$ has a removable discontinuity at $x=c$, we know that $\lim_{x \to c} f(x)$ exists but either $f(c)$ is undefined or $f(c) \neq \lim_{x \to c} f(x)$. For $h(x) = f(x)g(x)$ to be continuous at $x=c$, we need $\lim_{x \to c} h(x) = h(c)$. Using the product rule for limits: $\lim_{x \to c} h(x) = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x) = \lim_{x \to c} f(x) \cdot g(c)$ If $g(c) = 0$, then this limit equals $\lim_{x \to c} f(x) \cdot 0 = 0$. Also, $h(c) = f(c) \cdot g(c) = f(c) \cdot 0 = 0$ (assuming $f(c)$ is defined). Therefore, $\lim_{x \to c} h(x) = h(c) = 0$, making $h(x)$ continuous at $x=c$. This works regardless of the specific values of $f(c)$ or $\lim_{x \to c} f(x)$. Choice A sets an arbitrary relationship that doesn't ensure continuity of the product. Choice B ($g(c) = 1$) would make $\lim_{x \to c} h(x) = \lim_{x \to c} f(x)$, but $h(c) = f(c)$, and since $f$ is discontinuous, these likely won't be equal. Choice C creates the same problem as B since $f(c) \neq \lim_{x \to c} f(x)$ for removable discontinuities. Remember: multiplying by zero "kills" discontinuities. When you see products involving discontinuous functions, consider whether one factor being zero can force continuity.
2. For what value of the constant $k$ does the function $f(x) = \frac{x^2 + kx - 10}{x - 2}$ have a removable discontinuity at $x=2$?
   Answer: $k = 3$ For a rational function to have a removable discontinuity at $x=c$, the numerator must be zero at $x=c$, allowing the factor $(x-c)$ to be cancelled. Thus, the numerator $x^2 + kx - 10$ must be zero when $x=2$. Substituting $x=2$ gives: $(2)^2 + k(2) - 10 = 0 \Rightarrow 4 + 2k - 10 = 0 \Rightarrow 2k = 6 \Rightarrow k = 3$. When $k=3$, $f(x) = \frac{x^2+3x-10}{x-2} = \frac{(x+5)(x-2)}{x-2}$, which simplifies to $x+5$ for $x \neq 2$.
3. Which of the following statements accurately describes the function $f(x) = \frac{x^2 - 9}{|x+3|}$ at $x=-3$?
   Answer: It has a non-removable jump discontinuity because the left- and right-hand limits are not equal. To analyze the function at $x=-3$, we must evaluate the one-sided limits. For $x \to -3^+$, $x+3 > 0$, so $|x+3| = x+3$. The limit is $\lim_{x \to -3^+} \frac{(x-3)(x+3)}{x+3} = \lim_{x \to -3^+} (x-3) = -6$. For $x \to -3^-$, $x+3 < 0$, so $|x+3| = -(x+3)$. The limit is $\lim_{x \to -3^-} \frac{(x-3)(x+3)}{-(x+3)} = \lim_{x \to -3^-} -(x-3) = -(-3-3) = 6$. Since the left-hand limit (6) does not equal the right-hand limit (-6), the function has a non-removable jump discontinuity at $x=-3$.
4. A function is defined by $f(x) = \frac{\sqrt{x+3} - 2}{x - 1}$. To make this function continuous at $x=1$, what value should be assigned to $f(1)$?
   Answer: $\frac{1}{4}$ We need to find the limit of $f(x)$ as $x \to 1$. Direct substitution gives $\frac{0}{0}$, so we multiply the numerator and denominator by the conjugate of the numerator, $\sqrt{x+3} + 2$: $\lim_{x \to 1} \frac{\sqrt{x+3} - 2}{x - 1} \cdot \frac{\sqrt{x+3} + 2}{\sqrt{x+3} + 2} = \lim_{x \to 1} \frac{(x+3) - 4}{(x - 1)(\sqrt{x+3} + 2)} = \lim_{x \to 1} \frac{x - 1}{(x - 1)(\sqrt{x+3} + 2)}$. Canceling the $(x-1)$ terms gives $\lim_{x \to 1} \frac{1}{\sqrt{x+3} + 2} = \frac{1}{\sqrt{1+3} + 2} = \frac{1}{2+2} = \frac{1}{4}$.
5. The function $f(x) = \frac{x^3 + 27}{x^2 + 4x + 3}$ has a removable discontinuity. If a continuous function $g(x)$ is created by removing this discontinuity, what is the value of $g(-3)$?
   Answer: $-\frac{27}{2}$ The function has discontinuities where the denominator is zero: $x^2 + 4x + 3 = (x+3)(x+1) = 0$, so at $x=-3$ and $x=-1$. To check if the discontinuity at $x=-3$ is removable, we factor the numerator using the sum of cubes formula, $a^3+b^3 = (a+b)(a^2-ab+b^2)$. $x^3+27 = (x+3)(x^2 - 3x + 9)$. So, $f(x) = \frac{(x+3)(x^2 - 3x + 9)}{(x+3)(x+1)}$. The $(x+3)$ factor cancels, indicating a removable discontinuity at $x=-3$. The value of $g(-3)$ is the limit: $\lim_{x \to -3} \frac{x^2 - 3x + 9}{x+1} = \frac{(-3)^2 - 3(-3) + 9}{-3+1} = \frac{9+9+9}{-2} = \frac{27}{-2} = -\frac{27}{2}$.
6. The function $f(x) = \frac{x^p - 1}{x^q - 1}$ for integers $p, q > 0$ has a removable discontinuity at $x=1$. What is the value of the limit as $x \to 1$?
   Answer: $\frac{p}{q}$ When you encounter a function with a removable discontinuity, you're dealing with an indeterminate form that can be resolved by finding the limit. Here, both the numerator and denominator equal zero when $x = 1$, giving us the indeterminate form $\frac{0}{0}$. To find $\lim_{x \to 1} \frac{x^p - 1}{x^q - 1}$, you need to factor both expressions. Recall that $x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + x + 1)$. This means: - $x^p - 1 = (x-1)(x^{p-1} + x^{p-2} + \cdots + 1)$ - $x^q - 1 = (x-1)(x^{q-1} + x^{q-2} + \cdots + 1)$ After canceling the common factor $(x-1)$: $f(x) = \frac{x^{p-1} + x^{p-2} + \cdots + 1}{x^{q-1} + x^{q-2} + \cdots + 1}$ As $x \to 1$, the numerator becomes $1 + 1 + \cdots + 1$ ($p$ terms) $= p$, and the denominator becomes $1 + 1 + \cdots + 1$ ($q$ terms) $= q$. Therefore, the limit is $\frac{p}{q}$. Answer choice A) gives $1$, which would only be correct if $p = q$. Answer B) gives $p - q$, which incorrectly suggests subtraction rather than division. Answer C) is wrong because the limit can always be determined using this factorization method, regardless of the specific values of $p$ and $q$. Remember: when you see $x^n - 1$ in limits approaching $x = 1$, immediately think about factoring out $(x-1)$ to resolve the indeterminate form.
7. Consider the piecewise function $f(x) = \begin{cases} \frac{x-1}{1-\sqrt{x}} & x \in [0, 1) \\ a & x=1 \\ \frac{\sin(\pi x)}{x-1} & x > 1 \end{cases}$ Which statement is true regarding the continuity of $f(x)$ at $x=1$?
   Answer: The function has a jump discontinuity at $x=1$ for any value of $a$. For continuity at $x=1$, the left-hand limit, the right-hand limit, and the function value must all be equal. Left-hand limit: $\lim_{x \to 1^-} \frac{x-1}{1-\sqrt{x}} = \lim_{x \to 1^-} \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{-(\sqrt{x}-1)} = \lim_{x \to 1^-} -(\sqrt{x}+1) = -2$. Right-hand limit: $\lim_{x \to 1^+} \frac{\sin(\pi x)}{x-1}$. Let $u=x-1$, so $x=u+1$. As $x \to 1^+$, $u \to 0^+$. The limit becomes $\lim_{u \to 0^+} \frac{\sin(\pi(u+1))}{u} = \lim_{u \to 0^+} \frac{\sin(\pi u + \pi)}{u} = \lim_{u \to 0^+} \frac{-\sin(\pi u)}{u} = \lim_{u \to 0^+} \frac{-\sin(\pi u)}{\pi u} \cdot \pi = -1 \cdot \pi = -\pi$. Since the left-hand limit ($-2$) and the right-hand limit ($-\pi$) are not equal, the function has a non-removable jump discontinuity at $x=1$, regardless of the value of $a$.
8. Consider the function $f(x) = \frac{x^2 + kx - 10}{x-2}$. If the function has a removable discontinuity at $x=2$, what value must $f(2)$ be defined as to make the function continuous at that point?
   Answer: 7 For $f(x)$ to have a removable discontinuity at $x=2$, the numerator must be zero when $x=2$. This allows for the $(x-2)$ factor in the denominator to be cancelled. So, we set the numerator to zero: $(2)^2 + k(2) - 10 = 0$, which gives $4 + 2k - 10 = 0$, or $2k=6$, so $k=3$. Now, we find the limit of the function as $x$ approaches 2: $\lim_{x \to 2} \frac{x^2 + 3x - 10}{x-2} = \lim_{x \to 2} \frac{(x-2)(x+5)}{x-2} = \lim_{x \to 2} (x+5) = 2+5=7$. To remove the discontinuity, we must define $f(2)$ to be the value of the limit, which is 7.
9. Let $f(x) = \frac{\sqrt{4+x} - \sqrt{4-x}}{x}$. To remove the discontinuity at $x=0$, what value should be assigned to $f(0)$?
   Answer: $\frac{1}{2}$ When you see a function that's undefined at a point but appears to have a pattern, you're dealing with a removable discontinuity. The key is finding the limit as x approaches that point. Since $f(x) = \frac{\sqrt{4+x} - \sqrt{4-x}}{x}$ gives us $\frac{0}{0}$ when we substitute $x = 0$, we need to use algebraic manipulation to find $\lim_{x \to 0} f(x)$. The classic technique here is rationalizing by multiplying by the conjugate: $f(x) = \frac{\sqrt{4+x} - \sqrt{4-x}}{x} \cdot \frac{\sqrt{4+x} + \sqrt{4-x}}{\sqrt{4+x} + \sqrt{4-x}}$ Using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$: $f(x) = \frac{(4+x) - (4-x)}{x(\sqrt{4+x} + \sqrt{4-x})} = \frac{2x}{x(\sqrt{4+x} + \sqrt{4-x})} = \frac{2}{\sqrt{4+x} + \sqrt{4-x}}$ Now we can evaluate the limit: $\lim_{x \to 0} \frac{2}{\sqrt{4+0} + \sqrt{4-0}} = \frac{2}{2 + 2} = \frac{2}{4} = \frac{1}{2}$ Answer choice A ($0$) would result from incorrectly thinking the numerator approaching 0 makes the whole limit 0. Choice B ($\frac{1}{4}$) might come from errors in the algebraic manipulation. Choice C ($4$) could result from confusing the coefficient in the square roots with the final answer. Remember: when you see an indeterminate form involving square roots, rationalization is usually your best strategy to reveal the true limit.
10. Let $g(x) = \frac{x^2 - c^2}{|x|-c}$ for a constant $c > 0$. The function has removable discontinuities at $x=c$ and $x=-c$. Which of the following definitions for $h(x)$ creates a function that is continuous on $(-\infty, \infty)$?
    Answer: $h(x) = \begin{cases} g(x), & |x| \neq c \\ 2c, & |x| = c \end{cases}$ To make the function continuous, we must define $h(c)$ and $h(-c)$ to be equal to their respective limits. For $x=c$: As $x \to c$, we can assume $x>0$, so $|x|=x$. $\lim_{x \to c} \frac{x^2-c^2}{x-c} = \lim_{x \to c} \frac{(x-c)(x+c)}{x-c} = \lim_{x \to c} (x+c) = 2c$. For $x=-c$: As $x \to -c$, we can assume $x<0$, so $|x|=-x$. $\lim_{x \to -c} \frac{x^2-c^2}{-x-c} = \lim_{x \to -c} \frac{(x-c)(x+c)}{-(x+c)} = \lim_{x \to -c} -(x-c) = -(-c-c) = 2c$. Since both limits are $2c$, we must define the function to be $2c$ at both $x=c$ and $x=-c$ to remove the discontinuities.
11. A function $f(x)$ has a removable discontinuity at $x=c$. Which of the following statements MUST be true?
    Answer: $\lim_{x \to c} f(x)$ exists and is a finite number. When you encounter questions about removable discontinuities, think about what makes this type of discontinuity special compared to other types. A removable discontinuity means there's a "hole" in the function that could theoretically be "filled in" to make the function continuous at that point. For a discontinuity to be removable at $x = c$, the limit as $x$ approaches $c$ must exist and be finite. This is the defining characteristic. The function approaches the same finite value from both sides, but something prevents continuity at that exact point - either the function is undefined there, or it's defined but equals a different value than the limit. Choice B correctly states that $\lim_{x \to c} f(x)$ exists and is finite, which is exactly what defines a removable discontinuity. Choice A is wrong because while $f(c)$ is often undefined in removable discontinuities, it doesn't have to be. The function could be defined at $c$ but simply have the wrong value. Choice C describes a jump discontinuity, not a removable one. In removable discontinuities, the left and right limits are equal - that's why the overall limit exists. Choice D describes a vertical asymptote or infinite discontinuity, which is definitely not removable since you can't "fill in" an infinite gap. Remember: removable discontinuities are characterized by existing finite limits. If you can identify what value the function is "trying" to reach at the point, the discontinuity is likely removable.
12. The function $f(x) = \frac{e^{2x} - 1}{e^x - 1}$ has a removable discontinuity at $x=0$. What value should be assigned to $f(0)$ to make the function continuous?
    Answer: $2$ When you encounter a function with a removable discontinuity, you're dealing with a limit problem where the function appears undefined at a point but actually approaches a specific value. The key is finding what value the function approaches as x gets arbitrarily close to the problematic point. At $x = 0$, this function gives us $\frac{e^0 - 1}{e^0 - 1} = \frac{0}{0}$, an indeterminate form. To find the limiting value, we can factor the numerator using the difference of squares: $e^{2x} - 1 = (e^x)^2 - 1 = (e^x - 1)(e^x + 1)$. This gives us $f(x) = \frac{(e^x - 1)(e^x + 1)}{e^x - 1}$. For $x \neq 0$, we can cancel the $(e^x - 1)$ terms, leaving $f(x) = e^x + 1$. As $x$ approaches 0, this simplified expression approaches $e^0 + 1 = 1 + 1 = 2$. Therefore, setting $f(0) = 2$ makes the function continuous. Choice A (0) likely comes from incorrectly thinking the $\frac{0}{0}$ form equals zero. Choice C (1) might result from forgetting to add 1 after finding $e^0$. Choice D (e) could stem from misapplying exponential properties or incorrect algebraic manipulation. Study tip: When you see indeterminate forms like $\frac{0}{0}$, always look for algebraic simplification opportunities—factoring, rationalizing, or L'Hôpital's rule. The removable discontinuity's "repair value" is always the limit as x approaches the problematic point.
13. The function $f(x) = \frac{\sin(2x)}{x^2+x}$ has discontinuities at $x=0$ and $x=-1$. Which statement correctly describes how these discontinuities can be treated?
    Answer: Only the discontinuity at $x=0$ is removable; it is removed by defining $f(0)=2$. We analyze the limits at both points. At $x=0$: $\lim_{x \to 0} \frac{\sin(2x)}{x(x+1)} = \lim_{x \to 0} \frac{\sin(2x)}{2x} \cdot \frac{2}{x+1} = 1 \cdot \frac{2}{0+1} = 2$. Since the limit exists, the discontinuity at $x=0$ is removable. It can be removed by defining $f(0)=2$. At $x=-1$: $\lim_{x \to -1} \frac{\sin(2x)}{x(x+1)}$. The numerator approaches $\sin(-2)$, which is a non-zero number. The denominator approaches 0. This results in an infinite discontinuity (a vertical asymptote), which is not removable. Therefore, only the discontinuity at $x=0$ is removable.
14. Let $f(x) = p(x)/q(x)$ be a rational function where $p(x)$ and $q(x)$ are polynomials. If $f$ has a removable discontinuity at $x=c$, what can be concluded about the multiplicities of the root $x=c$ for $p(x)$ and $q(x)$? Let $m$ be the multiplicity of the root $x=c$ in $p(x)$ and $n$ be the multiplicity in $q(x)$.
    Answer: $m \geq n \geq 1$ For $f(x)$ to have a discontinuity at $x=c$, $q(c)$ must be 0, which means the multiplicity $n$ of the root $c$ in $q(x)$ must be at least 1 ($n \geq 1$). The function can be written as $f(x) = \frac{(x-c)^m P(x)}{(x-c)^n Q(x)}$ where $P(c), Q(c) \neq 0$. The limit as $x \to c$ is determined by the term $(x-c)^{m-n}$. For a removable discontinuity, this limit must be finite. This occurs if the exponent $m-n$ is greater than or equal to zero, i.e., $m \geq n$. If $m>n$, the limit is 0. If $m=n$, the limit is $P(c)/Q(c)$. Both are finite limits, corresponding to removable discontinuities. Therefore, the condition is $m \geq n \geq 1$.
15. Consider the function $f(x) = \frac{\ln(x)}{x-1}$. A new function $g(x)$ is defined to be continuous for all $x>0$ by removing the discontinuity of $f(x)$. What is the value of $g(1)$?
    Answer: $1$ The function $f(x)$ is undefined at $x=1$. To find the value that would make it continuous, we must calculate the limit as $x \to 1$. Direct substitution of $x=1$ into $f(x)$ results in the indeterminate form $\frac{\ln(1)}{1-1} = \frac{0}{0}$. We can apply L'Hôpital's Rule: $\lim_{x \to 1} \frac{\ln(x)}{x-1} = \lim_{x \to 1} \frac{\frac{d}{dx}(\ln(x))}{\frac{d}{dx}(x-1)} = \lim_{x \to 1} \frac{1/x}{1} = \frac{1/1}{1} = 1$. Therefore, defining $g(1)=1$ removes the discontinuity.
16. What value should be assigned to $f(0)$ to make the function $f(x) = \frac{1 - \cos^2(3x)}{x \sin(x)}$ continuous at $x=0$?
    Answer: 9 To find the value that removes the discontinuity at $x=0$, we need to evaluate the limit of $f(x)$ as $x \to 0$. First, use the Pythagorean identity $1 - \cos^2(\theta) = \sin^2(\theta)$. $f(x) = \frac{\sin^2(3x)}{x \sin(x)}$. Now, we evaluate the limit: $\lim_{x \to 0} \frac{\sin^2(3x)}{x \sin(x)} = \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot \frac{\sin(3x)}{x} \cdot \frac{3x}{x} \cdot \frac{x}{\sin(x)}$ is one way. A more direct way is: $\lim_{x \to 0} \frac{\sin^2(3x)}{(3x)^2} \cdot \frac{(3x)^2}{x \sin(x)} = \lim_{x \to 0} \left(\frac{\sin(3x)}{3x}\right)^2 \cdot \frac{9x^2}{x \sin(x)}$. Since $\lim_{u \to 0} \frac{\sin u}{u} = 1$, the first part of the product goes to $1^2=1$. The limit becomes: $\lim_{x \to 0} \frac{9x^2}{x \sin(x)} = \lim_{x \to 0} \frac{9x}{\sin(x)} = 9 \cdot \lim_{x \to 0} \frac{x}{\sin(x)} = 9 \cdot 1 = 9$. Therefore, to make the function continuous at $x=0$, we must define $f(0)=9$.
17. Let the function $f(x) = \frac{x^2 - 7x + 12}{x^2 - x - 6}$. Let $g(x)$ be a new function that is continuous for all real numbers and is identical to $f(x)$ except at $x=3$. What is the value of $g(3)$?
    Answer: $-\frac{1}{5}$ To find the value that makes the function continuous, we must evaluate the limit as $x$ approaches 3. First, factor the numerator and denominator: $f(x) = \frac{(x-3)(x-4)}{(x-3)(x+2)}$. For $x \neq 3$, we can cancel the $(x-3)$ terms, leaving $\frac{x-4}{x+2}$. The limit is then $\lim_{x \to 3} \frac{x-4}{x+2} = \frac{3-4}{3+2} = -\frac{1}{5}$. Therefore, defining $g(3) = -1/5$ will make the function continuous.
18. Let $f(x) = \frac{3x - \sin(x)}{2x + \sin(x)}$. A continuous function $g(x)$ is formed by defining $g(0) = \lim_{x \to 0} f(x)$. What is $g(0)$?
    Answer: $\frac{2}{3}$ To make the function continuous at $x=0$, we need to find the limit of $f(x)$ as $x \to 0$. Direct substitution gives $\frac{0}{0}$, an indeterminate form. We can use L'Hôpital's Rule. Differentiating the numerator and denominator with respect to $x$ gives: $\lim_{x \to 0} \frac{\frac{d}{dx}(3x - \sin(x))}{\frac{d}{dx}(2x + \sin(x))} = \lim_{x \to 0} \frac{3 - \cos(x)}{2 + \cos(x)}$. Now, substituting $x=0$ gives $\frac{3 - \cos(0)}{2 + \cos(0)} = \frac{3-1}{2+1} = \frac{2}{3}$. Alternatively, one could divide the numerator and denominator by $x$, which leads to $\lim_{x \to 0} \frac{3 - \sin(x)/x}{2 + \sin(x)/x} = \frac{3-1}{2+1} = \frac{2}{3}$. So, $g(0) = 2/3$.
19. The function $f(x) = \frac{1 - \cos^2(x)}{x^2}$ has a removable discontinuity at $x=0$. If $g(x)$ is the continuous extension of $f(x)$, what is $g(0)$?
    Answer: $1$ When you encounter a removable discontinuity problem, you're looking for a limit that exists but differs from (or is undefined at) the function's value at that point. The continuous extension assigns the limit value to make the function continuous. To find $g(0)$, you need to evaluate $\lim_{x \to 0} \frac{1 - \cos^2(x)}{x^2}$. The key insight is recognizing that $1 - \cos^2(x) = \sin^2(x)$ by the Pythagorean identity. This transforms your limit to: $\lim_{x \to 0} \frac{\sin^2(x)}{x^2} = \lim_{x \to 0} \left(\frac{\sin(x)}{x}\right)^2$ Since $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ (a fundamental limit), you get $1^2 = 1$. Therefore, $g(0) = 1$. Looking at the wrong answers: (A) $0$ might tempt you if you mistakenly think the numerator approaches zero faster than the denominator, but both approach zero at the same rate. (B) $\frac{1}{2}$ could result from incorrectly applying L'Hôpital's rule or misremembering trigonometric limits. (C) $2$ might come from algebraic errors when manipulating the trigonometric identities or confusing this with a different standard limit. Study tip: Master the fundamental trigonometric limits, especially $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ and $\lim_{x \to 0} \frac{1-\cos(x)}{x^2} = \frac{1}{2}$. Also remember that $\sin^2(x) + \cos^2(x) = 1$, so $1 - \cos^2(x) = \sin^2(x)$. These identities frequently appear in removable discontinuity problems.
20. The function $f(x) = \frac{e^{kx}-1}{ax+b}$ has a removable discontinuity at $x=0$. The continuous extension of $f(x)$ has a value of 1 at $x=0$, and $k=2$. What are the values of $a$ and $b$?
    Answer: $a=2, b=0$ When you encounter a function with a removable discontinuity, you're dealing with a situation where the function has an indeterminate form like $\frac{0}{0}$ at a point, but the limit exists. The key insight is that both the numerator and denominator must equal zero at the discontinuity point. Since $f(x) = \frac{e^{kx}-1}{ax+b}$ has a removable discontinuity at $x=0$, both the numerator and denominator must be zero when $x=0$. For the numerator: $e^{k \cdot 0} - 1 = e^0 - 1 = 1 - 1 = 0$ ✓. For the denominator: $a(0) + b = b$, so we need $b = 0$. Now we use the condition that the continuous extension has value 1 at $x=0$. This means $\lim_{x \to 0} f(x) = 1$. With $k=2$ and $b=0$, we have: $\lim_{x \to 0} \frac{e^{2x}-1}{ax} = 1$ Using L'Hôpital's rule (since we have $\frac{0}{0}$): $\lim_{x \to 0} \frac{2e^{2x}}{a} = \frac{2e^0}{a} = \frac{2}{a} = 1$ Therefore, $a = 2$. Looking at the wrong answers: A) gives $a=1$ (making the limit equal 2, not 1) and C) has $b=1$ (which wouldn't create the removable discontinuity since the denominator wouldn't be zero at $x=0$). D) gives $a=\frac{1}{2}$ (making the limit equal 4). The answer is B: $a=2, b=0$. Strategy tip: For removable discontinuities, always check that both numerator and denominator are zero at the problem point, then use L'Hôpital's rule or known limits to find the limiting value.

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