Types of Discontinuities - Calculus 1

1. Consider the function $f(x) = \frac{x^2 - 1}{\sin(\pi x)}$. Which of the following correctly describes the types of discontinuities for this function at integer values of $x$?
   Answer: Removable at $x=1$ and $x=-1$; infinite at all other integers. When analyzing discontinuities in rational functions, you need to identify where the function is undefined and then classify each discontinuity type. For $f(x) = \frac{x^2 - 1}{\sin(\pi x)}$, the function is undefined wherever $\sin(\pi x) = 0$, which occurs at all integer values of $x$. To classify each discontinuity, examine the behavior of both numerator and denominator at these points. At $x = 1$: the numerator equals $1^2 - 1 = 0$ and the denominator equals $\sin(\pi) = 0$. This creates a $\frac{0}{0}$ indeterminate form. Since $x^2 - 1 = (x-1)(x+1)$, you can factor out the $(x-1)$ term, and the limit exists as $x$ approaches 1. This makes the discontinuity removable. The same analysis applies at $x = -1$. At any other integer $n$ (where $n \neq \pm 1$), the numerator $n^2 - 1 \neq 0$ while the denominator still equals zero. This creates a $\frac{\text{nonzero}}{0}$ situation, causing the function to approach $\pm\infty$, making these infinite discontinuities. Choice A is wrong because not all integer discontinuities are removable—only $x = \pm 1$. Choice C incorrectly claims all discontinuities are infinite, missing the removable ones at $x = \pm 1$. Choice D mentions jump discontinuities, but rational functions don't typically have jump discontinuities; they have removable or infinite types. Study tip: For rational functions, check if both numerator and denominator equal zero at the same point—that's your clue for a potentially removable discontinuity.
2. Let $h(x) = g(f(x))$ where $f(x) = \frac{x^2-4}{x-2}$ and $g(x) = \frac{1}{x-4}$. Which of the following correctly describes the discontinuity of $h(x)$?
   Answer: An infinite discontinuity at $x=2$. When analyzing discontinuities in composite functions, you need to examine both where the inner function has issues and where the composition itself breaks down. First, let's simplify $f(x) = \frac{x^2-4}{x-2}$. Since $x^2-4 = (x+2)(x-2)$, we can cancel the common factor: $f(x) = \frac{(x+2)(x-2)}{x-2} = x+2$ for $x \neq 2$. However, $f(x)$ is undefined at $x = 2$, creating a removable discontinuity there. Now for the composite function $h(x) = g(f(x)) = g(x+2) = \frac{1}{(x+2)-4} = \frac{1}{x-2}$. This function is undefined when the denominator equals zero, which occurs at $x = 2$. Since $\lim_{x \to 2} \frac{1}{x-2} = \pm\infty$ (depending on direction), this creates an infinite discontinuity at $x = 2$. Let's check each option: Choice A incorrectly identifies the discontinuity type at $x = 2$ - it's infinite, not removable. Choice B suggests an infinite discontinuity at $x = 4$, but $h(4) = \frac{1}{4-2} = \frac{1}{2}$ is perfectly defined. Choice C also incorrectly places a discontinuity at $x = 4$ where none exists. Choice D correctly identifies an infinite discontinuity at $x = 2$. Study tip: With composite functions, always simplify the inner function first, then substitute into the outer function. Don't assume discontinuities from the original inner function automatically transfer to the composite - you must analyze the final simplified form.
3. Let $f(x) = \frac{2x^2+x+k}{x^2+x-2}$. If $\lim_{x \to -2} f(x)$ exists, what type of discontinuity does $f(x)$ have at $x=1$?
   Answer: Infinite The denominator is $x^2+x-2 = (x+2)(x-1)$, so the potential discontinuities are at $x=-2$ and $x=1$. The problem states that $\lim_{x \to -2} f(x)$ exists. For this limit to exist, the function must not have an infinite discontinuity at $x=-2$. This means the $(x+2)$ factor in the denominator must be cancelled by a factor in the numerator. Therefore, the numerator must be zero at $x=-2$. $2(-2)^2 + (-2) + k = 0 \implies 2(4) - 2 + k = 0 \implies 8 - 2 + k = 0 \implies k = -6$. Now the function is $f(x) = \frac{2x^2+x-6}{x^2+x-2}$. We can factor the numerator: $2x^2+x-6 = (2x-3)(x+2)$. So, $f(x) = \frac{(2x-3)(x+2)}{(x+2)(x-1)} = \frac{2x-3}{x-1}$ for $x \neq -2$. Now we analyze the discontinuity at $x=1$. We examine the limit $\lim_{x \to 1} \frac{2x-3}{x-1}$. The numerator approaches $2(1)-3 = -1$. The denominator approaches 0. A non-zero numerator and a zero denominator indicate an infinite discontinuity.
4. Let $f(x) = \frac{1-e^{x-2}}{x^2-4}$. Which of the following describes the type of discontinuity at $x=2$?
   Answer: Removable discontinuity The function has a potential discontinuity at $x=2$ because the denominator is zero. To classify it, we must evaluate the limit $\lim_{x \to 2} \frac{1-e^{x-2}}{x^2-4}$. This is an indeterminate form of type $0/0$. We can use L'Hôpital's Rule. The derivative of the numerator is $\frac{d}{dx}(1-e^{x-2}) = -e^{x-2}$. The derivative of the denominator is $\frac{d}{dx}(x^2-4) = 2x$. Applying L'Hôpital's Rule: $\lim_{x \to 2} \frac{-e^{x-2}}{2x} = \frac{-e^{2-2}}{2(2)} = \frac{-e^0}{4} = -\frac{1}{4}$. Since the limit exists and is finite, but the function is undefined at $x=2$, the discontinuity is removable.
5. A function $f(x)$ has a jump discontinuity at $x=c$, and a function $g(x)$ is continuous for all real numbers. If the product $h(x) = f(x)g(x)$ is continuous at $x=c$, which of the following MUST be true?
   Answer: $g(c) = 0$ Let the jump discontinuity at $x=c$ for $f(x)$ be characterized by $\lim_{x \to c^-} f(x) = L_1$ and $\lim_{x \to c^+} f(x) = L_2$, where $L_1 \neq L_2$. Since $g(x)$ is continuous, $\lim_{x \to c} g(x) = g(c)$. For $h(x) = f(x)g(x)$ to be continuous at $x=c$, its left-hand and right-hand limits must be equal. Left-hand limit of $h(x)$: $\lim_{x \to c^-} f(x)g(x) = L_1 \cdot g(c)$. Right-hand limit of $h(x)$: $\lim_{x \to c^+} f(x)g(x) = L_2 \cdot g(c)$. Setting them equal gives $L_1 \cdot g(c) = L_2 \cdot g(c)$, which simplifies to $(L_1 - L_2) \cdot g(c) = 0$. Since $f(x)$ has a jump discontinuity, $L_1 \neq L_2$, so $L_1 - L_2 \neq 0$. Therefore, we must have $g(c) = 0$.
6. Consider the function $f$ defined by $f(x) = \begin{cases} x^2 \cos(\frac{1}{x}) & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases}$ What type of discontinuity does $f(x)$ have at $x=0$?
   Answer: A removable discontinuity When analyzing discontinuities in piecewise functions, you need to examine what happens to the function's limit as you approach the point in question, then compare that to the function's actual value at that point. To determine the type of discontinuity at $x = 0$, let's find $\lim_{x \to 0} f(x)$. For $x \neq 0$, we have $f(x) = x^2 \cos(\frac{1}{x})$. Since $-1 \leq \cos(\frac{1}{x}) \leq 1$ for all $x \neq 0$, we can use the squeeze theorem: $-x^2 \leq x^2 \cos(\frac{1}{x}) \leq x^2$ As $x \to 0$, both $-x^2$ and $x^2$ approach 0, so $\lim_{x \to 0} x^2 \cos(\frac{1}{x}) = 0$. However, $f(0) = 1 \neq 0$. Since the limit exists but doesn't equal the function value, this is a removable discontinuity. Option A is incorrect because a jump discontinuity occurs when left and right limits exist but are unequal. Here, both one-sided limits equal 0. Option B is wrong because while $\cos(\frac{1}{x})$ oscillates wildly near zero, the $x^2$ factor forces the overall limit to exist—the oscillation doesn't prevent convergence. Option D is incorrect since $f(0) = 1$ while $\lim_{x \to 0} f(x) = 0$, creating a discontinuity. Remember: A discontinuity is removable when the limit exists but either the function isn't defined at that point or has the wrong value. You could "remove" the discontinuity by redefining $f(0) = 0$.
7. Let $f(x) = \frac{|x-a|}{x^2 - a^2}$ for a non-zero constant $a$. Which of the following correctly classifies the discontinuities of $f(x)$?
   Answer: A jump discontinuity at $x=a$ and an infinite discontinuity at $x=-a$. The function is $f(x) = \frac{|x-a|}{(x-a)(x+a)}$. We examine the points $x=a$ and $x=-a$. At $x=a$: We analyze the one-sided limits. For $x > a$, $|x-a| = x-a$, so $f(x) = \frac{x-a}{(x-a)(x+a)} = \frac{1}{x+a}$. Thus, $\lim_{x \to a^+} f(x) = \frac{1}{2a}$. For $x < a$, $|x-a| = -(x-a)$, so $f(x) = \frac{-(x-a)}{(x-a)(x+a)} = \frac{-1}{x+a}$. Thus, $\lim_{x \to a^-} f(x) = \frac{-1}{2a}$. Since $a \neq 0$, the left and right limits exist but are not equal, which defines a jump discontinuity. At $x=-a$: The limit is $\lim_{x \to -a} \frac{|x-a|}{(x-a)(x+a)}$. The numerator approaches $|-2a| = 2|a|$, which is a positive number. The denominator approaches 0. This results in an infinite limit, which defines an infinite discontinuity.
8. Consider the function $f(x) = \frac{\sin(2x-6)}{x^2-9}$. Which of the following correctly describes the function's discontinuities?
   Answer: A removable discontinuity at $x=3$ and an infinite discontinuity at $x=-3$. The denominator $x^2-9 = (x-3)(x+3)$ is zero at $x=3$ and $x=-3$. At $x=3$: We evaluate the limit $\lim_{x \to 3} \frac{\sin(2(x-3))}{(x-3)(x+3)}$. We can rewrite this as $\lim_{x \to 3} \frac{\sin(2(x-3))}{x-3} \cdot \frac{1}{x+3}$. Using the property $\lim_{u \to 0} \frac{\sin(ku)}{u} = k$, the first part of the limit is $\lim_{x \to 3} \frac{\sin(2(x-3))}{x-3} = 2$. The second part is $\lim_{x \to 3} \frac{1}{x+3} = \frac{1}{6}$. The overall limit is $2 \cdot \frac{1}{6} = \frac{1}{3}$. Since the limit exists, the discontinuity at $x=3$ is removable. At $x=-3$: We evaluate the limit $\lim_{x \to -3} \frac{\sin(2x-6)}{x^2-9}$. The numerator approaches $\sin(-12)$, which is a non-zero constant. The denominator approaches 0. A non-zero numerator with a denominator approaching zero indicates an infinite discontinuity.
9. Let $f(x)$ and $g(x)$ both have a jump discontinuity at $x=a$. Which statement about $h(x)=f(x)+g(x)$ is correct?
   Answer: $h(x)$ could be continuous, have a jump discontinuity, or have a removable discontinuity at $x=a$. Let $\lim_{x \to a^-} f(x) = L_1, \lim_{x \to a^+} f(x) = L_2$ with $L_1 \neq L_2$. Let $\lim_{x \to a^-} g(x) = M_1, \lim_{x \to a^+} g(x) = M_2$ with $M_1 \neq M_2$. For $h(x)=f(x)+g(x)$, the one-sided limits are $\lim_{x \to a^-} h(x) = L_1 + M_1$ and $\lim_{x \to a^+} h(x) = L_2 + M_2$. Case 1: If $L_1 + M_1 \neq L_2 + M_2$, then $h(x)$ has a jump discontinuity. Case 2: If $L_1 + M_1 = L_2 + M_2$ (the jumps cancel), then $\lim_{x \to a} h(x)$ exists. If $h(a)$ equals this limit, $h(x)$ is continuous. If $h(a)$ is undefined or different from the limit, $h(x)$ has a removable discontinuity. Therefore, all three types are possible.
10. Let $f(x) = \frac{1}{1-2^{\frac{1}{x}}}$. What kind of discontinuity does $f(x)$ have at $x=0$?
    Answer: Jump When analyzing discontinuities, you need to examine what happens to a function's behavior at points where it's undefined or exhibits sudden changes. For rational functions like this one, start by identifying where the function breaks down. Here, $f(x) = \frac{1}{1-2^{\frac{1}{x}}}$ is undefined when the denominator equals zero, which occurs when $1-2^{\frac{1}{x}} = 0$, or $2^{\frac{1}{x}} = 1$. Since $2^0 = 1$, this happens when $\frac{1}{x} = 0$, but this is impossible for any real number $x$. However, we need to examine the behavior as $x \to 0$. As $x \to 0^+$, we have $\frac{1}{x} \to +\infty$, so $2^{\frac{1}{x}} \to +\infty$, making $1-2^{\frac{1}{x}} \to -\infty$, and thus $f(x) \to 0^-$. As $x \to 0^-$, we have $\frac{1}{x} \to -\infty$, so $2^{\frac{1}{x}} \to 0$, making $1-2^{\frac{1}{x}} \to 1$, and thus $f(x) \to 1$. Since the left-hand limit (1) and right-hand limit (0) exist but are different, this is a jump discontinuity. A) Removable discontinuities occur when limits exist but don't equal the function value - not the case here since the limits differ. B) Oscillating discontinuities involve functions that oscillate wildly near the point - this function approaches distinct finite values. C) Infinite discontinuities occur when at least one one-sided limit is infinite - both our limits are finite. Study tip: For discontinuity classification, always check both one-sided limits first. Different finite limits = jump discontinuity.
11. The function $f(x)$ has a removable discontinuity at $x=a$ and $g(x)$ has an infinite discontinuity at $x=a$. Which of the following statements about the sum $h(x) = f(x) + g(x)$ at $x=a$ is always true?
    Answer: $h(x)$ has an infinite discontinuity. A removable discontinuity for $f(x)$ at $x=a$ means that $\lim_{x \to a} f(x) = L$ for some finite number L. An infinite discontinuity for $g(x)$ at $x=a$ means that $\lim_{x \to a} g(x) = \pm\infty$. To analyze the sum $h(x) = f(x) + g(x)$, we look at its limit: $\lim_{x \to a} h(x) = \lim_{x \to a} (f(x) + g(x))$. Using the limit laws for sums, this is $\lim_{x \to a} f(x) + \lim_{x \to a} g(x)$. This becomes $L + (\pm\infty)$. Adding a finite number $L$ to positive or negative infinity still results in positive or negative infinity. Therefore, $\lim_{x \to a} h(x) = \pm\infty$, which is the definition of an infinite discontinuity.
12. Describe the discontinuities of the function $f(x) = \frac{x^2 - x - 2}{x^3 - 2x^2}$.
    Answer: A removable discontinuity at $x=2$ and an infinite discontinuity at $x=0$. First, factor the numerator and denominator: $f(x) = \frac{(x-2)(x+1)}{x^2(x-2)}$. The function is undefined at $x=0$ and $x=2$.\ For $x \neq 2$, we can simplify the function to $f(x) = \frac{x+1}{x^2}$.\ To classify the discontinuity at $x=2$, we evaluate the limit: $\lim_{x\to 2} \frac{x+1}{x^2} = \frac{2+1}{2^2} = \frac{3}{4}$. Since the limit exists but $f(2)$ is undefined, there is a removable discontinuity at $x=2$.\ To classify the discontinuity at $x=0$, we evaluate the limit: $\lim_{x\to 0} \frac{x+1}{x^2}$. As $x \to 0$, the numerator approaches 1 and the denominator $x^2$ approaches 0 from the positive side. Thus, the limit is $+\infty$. This is an infinite discontinuity.
13. Let $f(x)$ be a function such that $\lim_{x \to 3} f(x) = 5$ and $f(3)$ is undefined. Let $g(x) = \frac{1}{f(x)-5}$. What type of discontinuity must $g(x)$ have at $x=3$?
    Answer: Infinite We are given that $\lim_{x \to 3} f(x) = 5$. This means as $x$ approaches 3, the value of $f(x)$ approaches 5. Now consider the function $g(x) = \frac{1}{f(x)-5}$. To classify the discontinuity at $x=3$, we examine the limit $\lim_{x \to 3} g(x)$. As $x \to 3$, the denominator $f(x)-5$ approaches $5-5=0$. The numerator is a constant 1. When the numerator is a non-zero constant and the denominator approaches zero, the limit goes to infinity or negative infinity. Therefore, $g(x)$ must have an infinite discontinuity at $x=3$. The fact that $f(3)$ is undefined means $g(3)$ is also undefined, but this does not change the nature of the infinite limit.
14. The function $f(x) = \frac{x^3 + x^2 - 12x}{x^2 - 9}$ has two discontinuities. Which of the following correctly classifies them?
    Answer: Removable at $x=3$, infinite at $x=-3$. First, factor the numerator and the denominator. Numerator: $x(x^2+x-12) = x(x+4)(x-3)$. Denominator: $x^2-9 = (x-3)(x+3)$. So, $f(x) = \frac{x(x+4)(x-3)}{(x-3)(x+3)}$. The discontinuities are at $x=3$ and $x=-3$. At $x=3$: The factor $(x-3)$ appears in both the numerator and the denominator. We can cancel it to find the limit: $\lim_{x \to 3} \frac{x(x+4)}{x+3} = \frac{3(3+4)}{3+3} = \frac{21}{6} = \frac{7}{2}$. Since the limit exists, the discontinuity at $x=3$ is removable. At $x=-3$: The factor $(x+3)$ is in the denominator but not the numerator (after cancellation). The limit is $\lim_{x \to -3} \frac{x(x+4)}{x+3}$. The numerator approaches $-3(-3+4) = -3$, while the denominator approaches 0. This is an infinite discontinuity.
15. The function $f(x) = \frac{(x-c)^2}{x^2 - c^2}$, where $c \neq 0$, has what types of discontinuities?
    Answer: A removable discontinuity at $x=c$ and an infinite discontinuity at $x=-c$. When analyzing discontinuities in rational functions, you need to examine points where the function is undefined and determine whether limits exist and are finite. First, let's identify where $f(x) = \frac{(x-c)^2}{x^2 - c^2}$ is undefined. The denominator $x^2 - c^2 = (x-c)(x+c) = 0$ when $x = c$ or $x = -c$, so the function has potential discontinuities at both points. At $x = c$: Since the numerator $(x-c)^2$ also equals zero here, we can simplify by factoring: $f(x) = \frac{(x-c)^2}{(x-c)(x+c)} = \frac{x-c}{x+c}$ (for $x \neq c$) Now $\lim_{x \to c} f(x) = \lim_{x \to c} \frac{x-c}{x+c} = \frac{0}{2c} = 0$. Since this limit exists and is finite, we have a removable discontinuity at $x = c$. At $x = -c$: Using our simplified form $\frac{x-c}{x+c}$, we get $\lim_{x \to -c} \frac{x-c}{x+c} = \frac{-2c}{0}$, which is infinite. This creates an infinite discontinuity. Choice A incorrectly swaps the types of discontinuities. Choice B wrongly claims both are infinite discontinuities—this misses the common factor cancellation at $x = c$. Choice C incorrectly claims both are removable—this ignores that the denominator still has $(x+c)$ as a factor after simplification. The correct answer is D: removable at $x = c$, infinite at $x = -c$. Study tip: Always factor both numerator and denominator completely to identify common factors that create removable discontinuities versus true vertical asymptotes.
16. Consider the function $f(x) = \begin{cases} \frac{x-1}{|x-1|} & \text{if } x \neq 1 \\ 0 & \text{if } x = 1 \end{cases}$. What is the nature of the discontinuity of the function $g(x) = xf(x)$ at $x=1$?
    Answer: Jump First, let's analyze $f(x)$. For $x>1$, $f(x)=1$. For $x<1$, $f(x)=-1$. This is a jump discontinuity at $x=1$. Now we analyze $g(x)=xf(x)$ at $x=1$. We check the one-sided limits: $\lim_{x \to 1^+} g(x) = \lim_{x \to 1^+} x \cdot (1) = 1$. $\lim_{x \to 1^-} g(x) = \lim_{x \to 1^-} x \cdot (-1) = -1$. Since the left-hand limit (-1) and the right-hand limit (1) exist but are not equal, the function $g(x)$ has a jump discontinuity at $x=1$. The value $g(1)=1 \cdot f(1) = 1 \cdot 0 = 0$ does not affect this classification.
17. The function $f(x) = \frac{3x^2 - 5x - 2}{x^2 - 4}$ has a removable discontinuity at $x=2$. What value should be assigned to $f(2)$ to make the function continuous at $x=2$?
    Answer: $7/4$ When you encounter a rational function with a removable discontinuity, you're dealing with a hole in the graph that can be "filled" by defining the function value at that point. The key is recognizing that both the numerator and denominator equal zero at the discontinuous point, creating an indeterminate form that can be resolved through algebraic manipulation. Since $f(x) = \frac{3x^2 - 5x - 2}{x^2 - 4}$ has a removable discontinuity at $x = 2$, both the numerator and denominator must equal zero when $x = 2$. Let's verify: the denominator $2^2 - 4 = 0$ ✓, and the numerator $3(4) - 5(2) - 2 = 12 - 10 - 2 = 0$ ✓. To find the correct value, factor both expressions. The denominator factors as $(x-2)(x+2)$. For the numerator, we need $3x^2 - 5x - 2 = (x-2)(3x+1)$. After canceling the common factor $(x-2)$: $\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{3x+1}{x+2} = \frac{3(2)+1}{2+2} = \frac{7}{4}$ Choice A is wrong because the discontinuity is removable, not essential. Choice B ($\frac{5}{4}$) likely comes from incorrectly factoring the numerator. Choice C ($0$) might result from thinking the limit equals zero since both numerator and denominator are zero at $x = 2$. Remember: for removable discontinuities, always factor and simplify first, then evaluate the limit. The correct value to assign is always this limit value.
18. Consider the function $f(x) = x - \lfloor x \rfloor$, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$. What type of discontinuity does $f(x)$ have at each integer $x=n$?
    Answer: A jump discontinuity Let's analyze the function at an arbitrary integer $x=n$. We need to check the one-sided limits.\ Left-hand limit: As $x \to n^-$, $x$ is slightly less than $n$, so $\lfloor x \rfloor = n-1$. Therefore, $\lim_{x\to n^-} (x - \lfloor x \rfloor) = n - (n-1) = 1$.\ Right-hand limit: As $x \to n^+$, $x$ is slightly greater than $n$, so $\lfloor x \rfloor = n$. Therefore, $\lim_{x\to n^+} (x - \lfloor x \rfloor) = n - n = 0$.\ Since the left-hand limit (1) and the right-hand limit (0) both exist but are unequal, the function has a jump discontinuity at every integer $n$.
19. Let $f(x) = \frac{\sqrt{x+6} - 3}{x^2-9}$. How can $f(x)$ be defined at $x=3$ to make it continuous at that point?
    Answer: $f(3) = \frac{1}{36}$ When you encounter a rational function that's undefined at a point due to both numerator and denominator being zero, you're dealing with a removable discontinuity. The key is finding the limit as x approaches that point. At $x = 3$, both the numerator $\sqrt{3+6} - 3 = 3 - 3 = 0$ and denominator $3^2 - 9 = 0$, giving the indeterminate form $\frac{0}{0}$. To find the removable discontinuity value, you need to evaluate $\lim_{x \to 3} f(x)$. The denominator factors as $x^2 - 9 = (x-3)(x+3)$. For the numerator, rationalize by multiplying by the conjugate: $\frac{\sqrt{x+6} - 3}{1} \cdot \frac{\sqrt{x+6} + 3}{\sqrt{x+6} + 3} = \frac{(x+6) - 9}{\sqrt{x+6} + 3} = \frac{x-3}{\sqrt{x+6} + 3}$ Now the function becomes: $f(x) = \frac{x-3}{(\sqrt{x+6} + 3)(x-3)(x+3)} = \frac{1}{(\sqrt{x+6} + 3)(x+3)}$ Taking the limit as $x \to 3$: $\lim_{x \to 3} f(x) = \frac{1}{(\sqrt{9} + 3)(6)} = \frac{1}{(3 + 3)(6)} = \frac{1}{36}$ Choice A gives $\frac{1}{18}$, likely from miscalculating the denominator. Choice B incorrectly assumes the discontinuity is non-removable when it's actually removable since we found a finite limit. Choice C of 6 appears to be the denominator value rather than the reciprocal. Remember: When you see $\frac{0}{0}$, always check if the discontinuity is removable by finding the limit through algebraic manipulation like factoring or rationalizing.
20. Let $f(x) = \frac{x^2 - 1}{|x|-1}$. Which of the following statements accurately describes the discontinuities of this function?
    Answer: Removable discontinuities at $x=1$ and $x=-1$. The function is undefined when $|x|-1 = 0$, which means at $x=1$ and $x=-1$. Let's analyze these points. For $x>0$, $|x|=x$, so $f(x) = \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1$ (for $x\neq 1$). The limit as $x \to 1$ is $\lim_{x \to 1} (x+1) = 2$. Since the limit exists but $f(1)$ is undefined, there is a removable discontinuity at $x=1$. For $x<0$, $|x|=-x$, so $f(x) = \frac{x^2-1}{-x-1} = \frac{(x-1)(x+1)}{-(x+1)} = -(x-1) = 1-x$ (for $x\neq -1$). The limit as $x \to -1$ is $\lim_{x \to -1} (1-x) = 1 - (-1) = 2$. Since the limit exists but $f(-1)$ is undefined, there is also a removable discontinuity at $x=-1$.

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