Infinite Limits & Vertical Asymptotes - Calculus 1

1. Given the function $f(x) = e^{1/(x-3)}$. Which of the following statements correctly describes the behavior of $f(x)$ at $x=3$?
   Answer: $\lim_{x \to 3^+} f(x) = \infty$, so $x=3$ is a vertical asymptote. To determine the behavior at $x=3$, we must evaluate the one-sided limits. As $x \to 3^+$, $x-3$ is a small positive number, so $1/(x-3) \to +\infty$. Thus, $\lim_{x \to 3^+} e^{1/(x-3)} = e^{+\infty} = +\infty$. As $x \to 3^-$, $x-3$ is a small negative number, so $1/(x-3) \to -\infty$. Thus, $\lim_{x \to 3^-} e^{1/(x-3)} = e^{-\infty} = 0$. A vertical asymptote exists if at least one of the one-sided limits is infinite. Since $\lim_{x \to 3^+} f(x) = \infty$, the line $x=3$ is a vertical asymptote.
2. A function $f(x)$ has a single vertical asymptote at $x=-1$. Furthermore, $\lim_{x \to -1^+} f(x) = -\infty$ and $\lim_{x \to -1^-} f(x) = -\infty$. Which of the following could be an equation for $f(x)$?
   Answer: $f(x) = \frac{-1}{(x+1)^2}$ The condition that the limit is the same ($-\infty$) from both the left and the right suggests that the term in the denominator that goes to zero should be squared, as this makes its sign positive regardless of the direction of approach. A denominator of $(x+1)^2$ approaches $0^+$ as $x \to -1$ from either side. For the limit to be $-\infty$, the numerator must be negative. In option C, $f(x) = \frac{-1}{(x+1)^2}$, the numerator is $-1$ and the denominator approaches $0^+$. Thus, $\lim_{x \to -1} f(x) = \frac{-1}{0^+} = -\infty$, which matches the given conditions.
3. Suppose $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = L$, where $L$ is a non-zero finite number. Let $h(x) = \frac{g(x)}{f(x)}$. Which of the following statements about $h(x)$ at $x=a$ must be true?
   Answer: $\lim_{x \to a} h(x) = 0$. We can evaluate the limit of $h(x)$ using the properties of limits. $\lim_{x \to a} h(x) = \lim_{x \to a} \frac{g(x)}{f(x)} = \frac{\lim_{x \to a} g(x)}{\lim_{x \to a} f(x)}$. Given that $\lim_{x \to a} g(x) = L$ (a finite non-zero number) and $\lim_{x \to a} f(x) = \infty$, the limit of the quotient is $\frac{L}{\infty}$, which is 0. Therefore, $\lim_{x \to a} h(x) = 0$. This means there is no vertical asymptote or jump discontinuity; rather, the function approaches a finite value.
4. What are the equations of all vertical asymptotes of the function $h(x) = \ln(|x^2-4|)$?
   Answer: $x = 2$ and $x = -2$ The natural logarithm function $\ln(u)$ has a vertical asymptote when its argument $u$ approaches 0 from the positive side. For the function $h(x) = \ln(|x^2-4|)$, the argument is $u = |x^2-4|$. We need to find the values of $x$ for which the argument approaches 0. This happens when $x^2-4 = 0$, which gives $x=2$ and $x=-2$. As $x$ approaches 2 from either side, $x^2-4$ approaches 0, so $|x^2-4|$ approaches $0^+$. Therefore, $\lim_{x \to 2} \ln(|x^2-4|) = -\infty$. This means $x=2$ is a vertical asymptote. Similarly, as $x$ approaches -2 from either side, $x^2-4$ approaches 0, so $|x^2-4|$ approaches $0^+$. Therefore, $\lim_{x \to -2} \ln(|x^2-4|) = -\infty$. This means $x=-2$ is a vertical asymptote. Thus, the function has two vertical asymptotes: $x=2$ and $x=-2$.
5. What are the equations of the vertical asymptotes of the function $f(x)$ defined as $f(x) = \begin{cases} \frac{1}{x+2} & \text{if } x < 0 \\ \frac{x-1}{x^2-1} & \text{if } x \ge 0 \end{cases}$?
   Answer: $x = -2$ only We analyze each piece of the function separately. For $x < 0$, the function is $f(x) = \frac{1}{x+2}$. A vertical asymptote occurs when the denominator is zero, which is at $x=-2$. Since $-2 < 0$, this is within the domain for this piece. Thus, $x=-2$ is a vertical asymptote. For $x \ge 0$, the function is $f(x) = \frac{x-1}{x^2-1} = \frac{x-1}{(x-1)(x+1)}$. Potential vertical asymptotes are at $x=1$ and $x=-1$. The value $x=-1$ is not in the domain $x \ge 0$, so we disregard it. For $x=1$, we check the limit: $\lim_{x \to 1} \frac{x-1}{(x-1)(x+1)} = \lim_{x \to 1} \frac{1}{x+1} = \frac{1}{2}$. Since the limit is a finite value, there is a hole at $x=1$, not a vertical asymptote. Finally, we check the boundary point $x=0$. The left-hand and right-hand limits are finite, so there is no asymptote there. Therefore, the only vertical asymptote is at $x=-2$.
6. How many vertical asymptotes does the function $f(x) = \sec(2x)$ have on the interval $[0, 2\pi]$?
   Answer: 4 The function $f(x) = \sec(2x)$ can be written as $1/\cos(2x)$. Vertical asymptotes occur when $\cos(2x) = 0$. The general solution for $\cos(u)=0$ is $u = \frac{\pi}{2} + n\pi$ for any integer $n$. In our case, $u=2x$, so $2x = \frac{\pi}{2} + n\pi$, which means $x = \frac{\pi}{4} + \frac{n\pi}{2}$. We need to find the number of integer values of $n$ for which $x$ is in the interval $[0, 2\pi]$. For $n=0$, $x = \pi/4$. For $n=1$, $x = \pi/4 + \pi/2 = 3\pi/4$. For $n=2$, $x = \pi/4 + \pi = 5\pi/4$. For $n=3$, $x = \pi/4 + 3\pi/2 = 7\pi/4$. For $n=4$, $x = \pi/4 + 2\pi = 9\pi/4$, which is outside the interval. Values of $n<0$ also give $x$ outside the interval. Thus, there are 4 vertical asymptotes.
7. Let $f(x) = \frac{x^2 - \pi^2}{\sin(x)}$. Which of the following is a vertical asymptote for $f(x)$?
   Answer: $x = 2\pi$ Vertical asymptotes can occur where the denominator is zero, which is $\sin(x) = 0$, or $x=n\pi$ for any integer $n$. We must check if the numerator is also zero at these points. The numerator is $x^2 - \pi^2 = (x-\pi)(x+\pi)$, which is zero at $x=\pi$ and $x=-\pi$. At $x=\pi$: The limit is of the form $0/0$. Using L'Hôpital's Rule: $\lim_{x \to \pi} \frac{2x}{\cos(x)} = \frac{2\pi}{-1} = -2\pi$. Since the limit is finite, $x=\pi$ is a hole. At $x=-\pi$: The limit is $0/0$. Using L'Hôpital's Rule: $\lim_{x \to -\pi} \frac{2x}{\cos(x)} = \frac{-2\pi}{-1} = 2\pi$. This is also a hole. At $x=2\pi$: The numerator is $(2\pi)^2 - \pi^2 = 3\pi^2 \neq 0$. The denominator is $\sin(2\pi)=0$. Since the limit is of the form $k/0$ with $k \neq 0$, the limit is infinite, and $x=2\pi$ is a vertical asymptote.
8. Let $f(x)$ be an odd function. If $\lim_{x \to 2^+} f(x) = -\infty$, which of the following limits must be true?
   Answer: $\lim_{x \to -2^-} f(x) = +\infty$ By definition, an odd function satisfies $f(-x) = -f(x)$. We are given $\lim_{x \to 2^+} f(x) = -\infty$. To find the limit as $x$ approaches $-2$, let $y = -x$. As $x \to 2^+$, $y \to -2^-$. We can write: $\lim_{y \to -2^-} f(y) = \lim_{x \to 2^+} f(-x)$. Using the odd function property, this becomes $\lim_{x \to 2^+} -f(x) = -\lim_{x \to 2^+} f(x)$. Substituting the given limit, we get $-(-\infty) = +\infty$. Therefore, $\lim_{x \to -2^-} f(x) = +\infty$.
9. Find all vertical asymptotes of $f(x) = \frac{x-3}{|x^2-9|}$.
   Answer: $x=-3$ only Potential vertical asymptotes occur where the denominator is zero, i.e., $|x^2-9|=0$, which means $x=3$ and $x=-3$. We must check the limits at these points. At $x=-3$: The numerator approaches $-6$ and the denominator approaches $0^+$. Thus, $\lim_{x \to -3} f(x) = \frac{-6}{0^+} = -\infty$. So, $x=-3$ is a vertical asymptote. At $x=3$: This is an indeterminate form. We rewrite $f(x) = \frac{x-3}{|x-3||x+3|}$. For $x \to 3^+$, $|x-3|=x-3$, so $f(x) = \frac{1}{|x+3|}$, and the limit is $1/6$. For $x \to 3^-$, $|x-3|=-(x-3)$, so $f(x) = \frac{-1}{|x+3|}$, and the limit is $-1/6$. Since both one-sided limits are finite, there is no vertical asymptote at $x=3$.
10. A function $f(x)$ is defined for all $x \neq c$. If $\lim_{x \to c} (f(x))^2 = L$, where $L$ is a finite positive number, what must be true about the graph of $f(x)$ at $x=c$?
    Answer: The function does not have a vertical asymptote at $x=c$. For the line $x=c$ to be a vertical asymptote, $\lim_{x \to c} |f(x)|$ must be $\infty$. If this were the case, then $\lim_{x \to c} (f(x))^2$ would also be $\infty$. However, we are given that this limit is a finite positive number $L$. This implies that $\lim_{x \to c} |f(x)| = \sqrt{L}$, which is a finite value. Since the limit of the absolute value of the function is finite, the function cannot have a vertical asymptote at $x=c$. It could have a removable discontinuity or a jump discontinuity, but we cannot be certain which without more information.
11. Consider the function $f(x) = \begin{cases} \frac{1}{x} & \text{if } x < 0 \\ x^2+1 & \text{if } x \ge 0 \end{cases}$. Which statement is true?
    Answer: $x=0$ is a vertical asymptote because $\lim_{x \to 0^-} f(x) = -\infty$. When analyzing piecewise functions for vertical asymptotes, you need to examine the behavior near potential discontinuities by checking one-sided limits, especially where the function might blow up to infinity. For this piecewise function, $x = 0$ is the boundary point where the definition changes. To determine if there's a vertical asymptote, check what happens as you approach from each side. From the left ($x \to 0^-$), you're using $f(x) = \frac{1}{x}$. As $x$ approaches 0 from negative values, $\frac{1}{x}$ becomes increasingly negative, approaching $-\infty$. This unbounded behavior is exactly what defines a vertical asymptote. Choice A is correct because it accurately identifies that $\lim_{x \to 0^-} f(x) = -\infty$, which creates a vertical asymptote at $x = 0$. Choice B is wrong because having a defined function value doesn't prevent vertical asymptotes. While $f(0) = 0^2 + 1 = 1$ is indeed defined, vertical asymptotes are about limit behavior, not whether the function is defined at that point. Choice C misses the key issue. While $\lim_{x \to 0^+} f(x) = 1$ is finite, vertical asymptotes only require one-sided limits to be infinite. The left-hand limit going to $-\infty$ is sufficient. Choice D is false because continuity requires both one-sided limits to exist, be finite, and equal $f(0)$. Since the left-hand limit is $-\infty$, the function has a discontinuity. Study tip: For vertical asymptotes in piecewise functions, check both one-sided limits at boundary points—you only need one side to blow up to infinity.
12. Suppose $\lim_{x \to 0} f(x) = \infty$. Which of the following limits must also be infinite? I. $\lim_{x \to 0} (f(x) + \sin(x))$ II. $\lim_{x \to 0} (x \cdot f(x))$ III. $\lim_{x \to 0} (f(x))^2$
    Answer: I and III only I. As $x \to 0$, $f(x) \to \infty$. The function $\sin(x)$ is bounded between -1 and 1. Adding a bounded function to a function that grows to infinity will still result in a function that grows to infinity. So, $\lim_{x \to 0} (f(x) + \sin(x)) = \infty$. Statement I is true. II. This limit is of the indeterminate form $0 \cdot \infty$. Its value is not necessarily infinite. For example, if $f(x) = 1/x$, the limit is 1. If $f(x) = 1/x^2$, the limit is infinite. Since it is not guaranteed to be infinite, statement II is false. III. If $f(x) \to \infty$, then squaring it will also result in a limit of $\infty$. So, $\lim_{x \to 0} (f(x))^2 = \infty$. Statement III is true.
13. For what value of $c$ does the function $f(x) = \frac{x^2+4}{x^2+2cx+c^2}$ have a vertical asymptote at $x=3$?
    Answer: $c = -3$ When you encounter a rational function and need to find where it has a vertical asymptote, you're looking for values that make the denominator zero while keeping the numerator non-zero. This creates the undefined behavior that produces a vertical asymptote. For $f(x) = \frac{x^2+4}{x^2+2cx+c^2}$ to have a vertical asymptote at $x=3$, the denominator must equal zero when $x=3$. Let's substitute: $3^2 + 2c(3) + c^2 = 0$, which gives us $9 + 6c + c^2 = 0$. Rearranging: $c^2 + 6c + 9 = 0$. This factors as $(c+3)^2 = 0$, so $c = -3$. Let's verify: when $c = -3$, the denominator becomes $x^2 - 6x + 9 = (x-3)^2$. At $x=3$, this equals zero. The numerator $3^2 + 4 = 13 \neq 0$, confirming we get a vertical asymptote. Choice A ($c = 3$) gives denominator $x^2 + 6x + 9 = (x+3)^2$, which is zero at $x = -3$, not $x = 3$. Choice C ($c = 2$) gives $x^2 + 4x + 4 = (x+2)^2$, creating an asymptote at $x = -2$. Choice B is incorrect because we've found that $c = -3$ works perfectly. Remember: to find vertical asymptotes, set the denominator equal to zero and solve. Always check that the numerator isn't also zero at that point, which would create a removable discontinuity instead.
14. Which of the following statements provides a necessary and sufficient condition for the line $x=c$ to be a vertical asymptote of the function $f(x)$?
    Answer: At least one of the one-sided limits, $\lim_{x \to c^+} f(x)$ or $\lim_{x \to c^-} f(x)$, is $\infty$ or $-\infty$. The definition of a vertical asymptote is based on the behavior of the function's limit. The line $x=c$ is a vertical asymptote if and only if the limit of the function as $x$ approaches $c$ from the left or the right is positive or negative infinity. This is precisely what statement C says. Statement A is sufficient for rational functions but not necessary for all functions (e.g., $\ln(x)$ has a VA at $x=0$). Statement B is not sufficient, as it could be a hole. Statement D is not sufficient, as a jump discontinuity is non-removable but not a vertical asymptote.
15. Let P be the statement "The function $f$ has a vertical asymptote at $x=c$" and Q be the statement "$f(c)$ is undefined." Which of the following correctly describes the logical relationship between P and Q?
    Answer: Neither statement is a necessary or sufficient condition for the other. P does not imply Q: A function can have a vertical asymptote at $x=c$ even if $f(c)$ is defined. For example, $f(x) = \begin{cases} 1/x & x \neq 0 \\ 5 & x=0 \end{cases}$ has a vertical asymptote at $x=0$, but $f(0)=5$. Q does not imply P: A function can be undefined at $x=c$ without having a vertical asymptote. For example, $f(x) = \frac{x^2-1}{x-1}$ is undefined at $x=1$, but it has a hole (a removable discontinuity), not a vertical asymptote. Therefore, neither statement implies the other.
16. For which of the following functions does $\lim_{x \to 2} f(x)$ exist and is finite, even though the expression for $f(x)$ has a denominator that is zero at $x=2$?
    Answer: $f(x) = \frac{\sin(x-2)}{x^2-4}$ When you encounter a rational function where the denominator equals zero at a specific point, you're dealing with a potential removable discontinuity (hole) versus a vertical asymptote. The key is determining whether the numerator also approaches zero at that point, creating an indeterminate form that might have a finite limit. Let's examine each function at $x = 2$: For option D, $f(x) = \frac{\sin(x-2)}{x^2-4}$, both the numerator and denominator approach zero as $x \to 2$. Since $x^2-4 = (x-2)(x+2)$, we can rewrite this as $\frac{\sin(x-2)}{(x-2)(x+2)}$. Using the fundamental limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$, we get: $\lim_{x \to 2} \frac{\sin(x-2)}{x^2-4} = \lim_{x \to 2} \frac{\sin(x-2)}{x-2} \cdot \frac{1}{x+2} = 1 \cdot \frac{1}{4} = \frac{1}{4}$ Option A has $f(x) = \frac{x^2}{x-2}$. As $x \to 2$, the numerator approaches 4 while the denominator approaches 0, creating a vertical asymptote with an infinite limit. Option B has $f(x) = \frac{e^x}{(x-2)^2}$. The numerator approaches $e^2$ while the denominator approaches 0, again yielding an infinite limit. Option C has $f(x) = \frac{x+2}{x^2-4} = \frac{x+2}{(x-2)(x+2)} = \frac{1}{x-2}$ (after canceling $x+2$), which still has a vertical asymptote at $x = 2$. Remember: Look for cases where both numerator and denominator approach zero simultaneously—these often yield finite limits after applying L'Hôpital's rule or algebraic manipulation.
17. What are the equations of the vertical asymptotes for the function $f(x) = \ln(x^2 - x - 6)$?
    Answer: $x=-2$ and $x=3$ The natural logarithm function $\ln(u)$ has a vertical asymptote where its argument $u$ approaches 0 from the positive side. For $f(x) = \ln(x^2 - x - 6)$, the argument is $u = x^2 - x - 6$. We set the argument to zero to find potential asymptotes: $x^2 - x - 6 = 0 \implies (x-3)(x+2) = 0$, which gives $x=3$ and $x=-2$. The domain of the function requires $x^2-x-6 > 0$, which is true for $x>3$ or $x<-2$. As $x \to 3^+$, the argument approaches $0^+$, so $\lim_{x \to 3^+} f(x) = -\infty$. As $x \to -2^-$, the argument approaches $0^+$, so $\lim_{x \to -2^-} f(x) = -\infty$. Thus, both $x=3$ and $x=-2$ are vertical asymptotes.
18. The function $f(x) = \frac{x^2+ax-6}{x^2-4}$ has a removable discontinuity at $x=2$. What is the equation of its vertical asymptote?
    Answer: $x = -2$ For a rational function to have a removable discontinuity at $x=2$, $(x-2)$ must be a factor of both the numerator and the denominator. This means the numerator must be zero when $x=2$. So, $2^2 + a(2) - 6 = 0 \implies 4 + 2a - 6 = 0 \implies 2a = 2 \implies a=1$. Substituting $a=1$ into the function gives $f(x) = \frac{x^2+x-6}{x^2-4} = \frac{(x+3)(x-2)}{(x+2)(x-2)}$. After canceling the $(x-2)$ terms, the simplified function is $\frac{x+3}{x+2}$. The vertical asymptote occurs where the denominator of the simplified function is zero, which is $x+2=0$, or $x=-2$.
19. Let $f(x) = \frac{P(x)}{Q(x)}$ be a rational function where $P(x)$ and $Q(x)$ are non-zero polynomials. Which of the following statements is not always true?
    Answer: If $Q(a)=0$ and $P(a)=0$, then $f(x)$ has a hole at $x=a$. Let's analyze each statement. A) This is the fundamental condition for a vertical asymptote in a rational function. A non-zero numerator divided by a denominator approaching zero results in an infinite limit. This statement is always true. B) For a rational function to have an infinite limit at a finite value $x=a$, the denominator must approach zero. Since $Q(x)$ is a polynomial, this means $Q(a)=0$. This statement is always true. C) If $Q(a)=0$ and $P(a)=0$, it means $(x-a)$ is a factor of both polynomials. This leads to an indeterminate form $\frac{0}{0}$. While this often results in a hole (removable discontinuity), it can also result in a vertical asymptote if the multiplicity of the root $a$ is higher in the denominator $Q(x)$ than in the numerator $P(x)$. For example, if $f(x) = \frac{(x-a)}{(x-a)^2} = \frac{1}{x-a}$, there is a vertical asymptote at $x=a$. Thus, this statement is not always true. D) The vertical asymptotes of a rational function can only occur at the real roots of the denominator polynomial $Q(x)$. A polynomial of degree $n$ has at most $n$ real roots. Therefore, the number of vertical asymptotes is at most the degree of $Q(x)$. This statement is always true.
20. The domain of the function $f(x) = \sqrt{\frac{x}{x-2}}$ is $(-\infty, 0] \cup (2, \infty)$. Based on this domain, what can be concluded about vertical asymptotes?
    Answer: $x=2$ is a vertical asymptote, but $x=0$ is not. When analyzing functions with square roots and fractions, you need to consider both domain restrictions and vertical asymptote behavior. A vertical asymptote occurs where a function approaches infinity as x approaches a specific value, but crucially, the function must be defined on at least one side of that point. Let's examine what happens at the critical points. At $x = 0$, the expression $\frac{x}{x-2} = \frac{0}{-2} = 0$, so $f(0) = \sqrt{0} = 0$. The function actually has a value here, so there's no vertical asymptote at $x = 0$. At $x = 2$, the denominator becomes zero, making $\frac{x}{x-2}$ undefined. More importantly, looking at the given domain $(-\infty, 0] \cup (2, \infty)$, notice that the function is defined for values immediately to the right of $x = 2$ (like $x = 2.1$). As $x$ approaches 2 from the right, $\frac{x}{x-2}$ approaches $+\infty$, so $f(x) = \sqrt{\frac{x}{x-2}}$ also approaches $+\infty$. This creates a vertical asymptote at $x = 2$. Choice A is wrong because $x = 0$ is not a vertical asymptote—the function equals zero there. Choice B is incorrect since we do have a vertical asymptote at $x = 2$. Choice C reverses the correct relationship between the two points. Study tip: For vertical asymptotes, don't just look for where expressions are undefined—check whether the function approaches infinity near those points and whether the domain includes values on at least one side of the potential asymptote.

Start this quiz