Connecting Infinite Limits and Vertical Asymptotes - AP Calculus BC

Let $p(x)=\dfrac{2}{(x+1)^2}$ . Which statement about the limit as $x\to-1$ is correct?
Answer: There is a vertical asymptote at $x=-1$ and $\lim_{x\to-1} p(x)=\infty$ . This question tests your ability to connect infinite limits with vertical asymptotes in functions with even-powered denominators. For p(x) = 2/ $(x+1)^2$ , the denominator approaches zero at x=-1, and since it's squared, it approaches from the positive side regardless of direction. Thus, the function approaches positive infinity from both sides as x nears -1. The numerator is constant and positive, reinforcing the infinite limit. Choice E is tempting but fails because the limit is infinite, not zero; zero would require the numerator to approach zero as well. Always consider the sign and power of terms near potential asymptotes to determine limit directions.
For $u(x)=\dfrac{x-1}{\sqrt{x-1}}$ with domain $x>1$ , which statement describes $\lim_{x\to1^+} u(x)$ ?
Answer: $\lim_{x\to1^+} u(x)=0$ , so $x=1$ is not a vertical asymptote This problem tests simplifying algebraic expressions before taking limits. For $u(x)=\frac{x-1}{\sqrt{x-1}}$ with $x>1$ , we can rewrite this as $u(x)=\frac{x-1}{(x-1)^{1/2}}=(x-1)^{1-1/2}=(x-1)^{1/2}=\sqrt{x-1}$ . As $x\to1^+$ , we have $\sqrt{x-1}\to\sqrt{0^+}=0$ . Since the limit is finite (zero), there is no vertical asymptote at $x=1$ . Choice A incorrectly assumes an infinite limit without simplifying. Always simplify expressions algebraically before evaluating limits to avoid missing cancellations.
For $u(x)=\dfrac{x-1}{(x-1)^2}$ , which statement correctly describes the vertical asymptote behavior at $x=1$ ?
Answer: Vertical asymptote at $x=1$ with $\lim_{x\to 1^-}u(x)=-\infty$ and $\lim_{x\to 1^+}u(x)=+\infty$ This question tests understanding of vertical asymptotes when factors appear with different powers in numerator and denominator. The function u(x) = (x-1)/(x-1)² simplifies to u(x) = 1/(x-1) for x ≠ 1, since one factor of (x-1) cancels. After simplification, as x approaches 1 from the left, (x-1) approaches 0 through negative values, making 1/(x-1) approach -∞; as x approaches 1 from the right, (x-1) approaches 0 through positive values, making 1/(x-1) approach +∞. This creates a vertical asymptote at x = 1 with the described limit behavior. Choice E incorrectly suggests no vertical asymptote due to cancellation, but the cancellation is only partial. When the denominator has a higher power of a factor than the numerator, a vertical asymptote still occurs after cancellation.
For $u(x)=\dfrac{5}{x}-2$ , which statement correctly describes the vertical asymptote and limit at it?
Answer: Vertical asymptote at $x=0$ and $\lim_{x\to0^+} u(x)=\infty$ , $\lim_{x\to0^-} u(x)=-\infty$ . This question tests your ability to connect infinite limits with vertical asymptotes in transformed reciprocal functions. For u(x) = 5/x - 2, the 5/x term causes infinite behavior at x=0: positive infinity from the right and negative infinity from the left. Subtracting 2 shifts the graph vertically but doesn't prevent the limits from being infinite. Thus, there's a vertical asymptote at x=0 with opposite-signed infinite limits. Choice D is tempting but fails because the shift affects horizontal asymptotes, not the vertical one from the 1/x core. Always isolate the term causing division by zero and assess its infinite limits, regardless of added constants.
For $h(x)=\dfrac{x^2-9}{x-3}$ , which statement correctly describes the behavior near $x=3$ ?
Answer: $\lim_{x\to 3} h(x)$ exists and is finite. This question tests whether you can recognize when a rational function has a removable discontinuity rather than a vertical asymptote. For $h(x)=\frac{x^2-9}{x-3}$ , both numerator and denominator equal zero when $x=3$ , creating an indeterminate form $\frac{0}{0}$ . We can factor the numerator as $(x-3)(x+3)$ and simplify: $h(x)=\frac{(x-3)(x+3)}{x-3}=x+3$ for $x\neq 3$ . Taking the limit as $x\to 3$ , we get $\lim_{x\to 3} h(x)=3+3=6$ , which is finite. Therefore, there's a hole (removable discontinuity) at $x=3$ , not a vertical asymptote. Choice A incorrectly claims there's a vertical asymptote at $x=3$ , missing that the common factor cancels. When both numerator and denominator share a zero, always factor and simplify before determining asymptote behavior.
For $f(x)=\dfrac{3x+1}{x^2-4}$ , which statement correctly describes the vertical asymptote behavior near $x=2$ ?
Answer: $\lim_{x\to2^-} f(x)=-\infty$ and $\lim_{x\to2^+} f(x)=+\infty$ , so $x=2$ is a vertical asymptote This question tests the skill of using infinite limits to identify vertical asymptotes in rational functions. The function f(x) = (3x+1)/(x²-4) has a denominator that factors to (x-2)(x+2), which is zero at x=2 while the numerator is 7, indicating a potential vertical asymptote. As x approaches 2 from the left, the function goes to negative infinity because the denominator is negative and the numerator positive, creating a large negative value. From the right, it approaches positive infinity as both are positive, confirming the asymptote at x=2. A tempting distractor might suggest a removable discontinuity because the limit appears as 7/0, but this undefined form actually signals infinite behavior rather than a finite limit after simplification. To generally identify vertical asymptotes, evaluate one-sided limits where the denominator is zero and the numerator is not, checking if they approach infinity.
For $h(x)=\dfrac{2x-4}{x^2-4}$ , which statement about asymptotes at $x=2$ is correct?
Answer: $\lim_{x\to2} h(x)=\dfrac{1}{2}$ , so $x=2$ is not a vertical asymptote This problem tests recognizing when a removable discontinuity prevents a vertical asymptote. For $h(x)=\frac{2x-4}{x^2-4}=\frac{2(x-2)}{(x-2)(x+2)}$ , we can cancel the common factor $(x-2)$ to get $h(x)=\frac{2}{x+2}$ for $x\neq2$ . Taking the limit as $x\to2$ , we get $\lim_{x\to2} h(x)=\frac{2}{2+2}=\frac{1}{2}$ , which is finite. Since the limit exists and is finite, there is no vertical asymptote at $x=2$ (just a hole). Choice A incorrectly assumes an infinite limit without checking for cancellation. Always factor and simplify rational functions before determining asymptotes.
Let $t(x)=\dfrac{\sin(x)}{x-\pi}$ . Which statement best describes the behavior as $x\to\pi$ ?
Answer: There is a removable discontinuity at $x=\pi$ and $\lim_{x\to\pi} t(x)=-1$ . This question tests your ability to connect infinite limits with vertical asymptotes in trigonometric rational functions. For t(x) = sin(x)/(x-π), as x approaches π, it's 0/0 indeterminate, but simplifying or using L'Hôpital's rule gives limit cos(π) = -1, which is finite. Thus, there's no infinite behavior, indicating a removable discontinuity. The sine function's continuity doesn't prevent the hole from the zero denominator. Choice A is tempting but fails because the numerator also approaches zero, making the limit finite after analysis. Always check if both numerator and denominator approach zero; if so, simplify to see if the limit is finite, ruling out asymptotes.
For $s(x)=\dfrac{1}{(x+2)^3}$ , which statement correctly describes the asymptote behavior as $x\to-2$ ?
Answer: $\lim_{x\to-2^-} s(x)=-\infty$ and $\lim_{x\to-2^+} s(x)=\infty$ This problem tests understanding infinite limits with odd powers in the denominator. For $s(x)=\frac{1}{(x+2)^3}$ , as $x\to-2$ , the denominator approaches zero. Since the power is odd, the sign matters: from the left ( $x<-2$ ), we have $(x+2)<0$ , so $(x+2)^3<0$ , giving $\lim_{x\to-2^-} s(x)=\frac{1}{0^-}=-\infty$ . From the right ( $x>-2$ ), we have $(x+2)>0$ , so $(x+2)^3>0$ , giving $\lim_{x\to-2^+} s(x)=\frac{1}{0^+}=+\infty$ . These opposite infinite limits confirm $x=-2$ is a vertical asymptote. Choice C incorrectly claims both limits are positive. For odd powers, one-sided limits have opposite signs at vertical asymptotes.
For $f(x)=\dfrac{3x-1}{(x+2)(x-5)}$ , which statement correctly describes the vertical asymptote behavior?
Answer: Vertical asymptotes at $x=-2$ and $x=5$ This question tests your ability to identify vertical asymptotes by analyzing infinite limits of rational functions. The function f(x) = (3x-1)/[(x+2)(x-5)] has its denominator equal to zero when x = -2 and x = 5, while the numerator is non-zero at these values. As x approaches -2 or 5, the denominator approaches 0 while the numerator approaches finite non-zero values, causing the function to approach ±∞. This creates vertical asymptotes at x = -2 and x = 5. Choice B incorrectly identifies these as horizontal asymptotes, which would be lines of the form y = k, not x = k. When a rational function's denominator has zeros that don't cancel with the numerator, vertical asymptotes occur at those x-values.
For $q(x)=\dfrac{2}{x-4}+1$ , which statement correctly describes the vertical asymptote?
Answer: There is a vertical asymptote at $x=4$ . This question tests your ability to identify vertical asymptotes in transformed rational functions. For $q(x)=\frac{2}{x-4}+1$ , the rational part $\frac{2}{x-4}$ has a denominator that equals zero when $x=4$ . As $x\to 4^-$ , we have $(x-4)\to 0^-$ , so $\frac{2}{x-4}\to -\infty$ , making $q(x)\to -\infty$ . As $x\to 4^+$ , we have $(x-4)\to 0^+$ , so $\frac{2}{x-4}\to +\infty$ , making $q(x)\to +\infty$ . This confirms a vertical asymptote at $x=4$ . Choice D incorrectly calls this a horizontal asymptote, confusing the direction of the asymptote line. Remember that vertical asymptotes are vertical lines $x=a$ where the function has infinite limits, while horizontal asymptotes are horizontal lines $y=b$ describing end behavior.
For $t(x)=\dfrac{(x-2)^2}{(x-2)(x+3)}$ , which statement correctly identifies any vertical asymptote?
Answer: $\lim_{x\to-3} t(x)=\pm\infty$ , so $x=-3$ is a vertical asymptote This problem tests identifying vertical asymptotes via infinite limits after simplifying rational functions. For t(x) = (x-2)² / [(x-2)(x+3)], it simplifies to (x-2)/(x+3) for x ≠ 2, with a hole at x=2 and denominator zero at x=-3. As x approaches -3, the simplified function goes to positive infinity from the left and negative infinity from the right. These infinite limits confirm a vertical asymptote at x=-3. A distractor might focus on x=2 due to the original form, but simplification reveals a removable discontinuity there instead. To find asymptotes, simplify the function first, then check limits at points where the simplified denominator is zero.
Let $v(x)=\dfrac{x^2-1}{x-1}$ . Which statement correctly describes the limit and asymptote behavior at $x=1$ ?
Answer: $\lim_{x\to1} v(x)=2$ , so $x=1$ is not a vertical asymptote This question tests recognizing removable discontinuities in factored rational functions. For $v(x)=\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}$ , we can cancel the common factor $(x-1)$ to get $v(x)=x+1$ for $x\neq1$ . Taking the limit: $\lim_{x\to1} v(x)=\lim_{x\to1} (x+1)=1+1=2$ . Since the limit exists and equals 2 (finite), there is no vertical asymptote at $x=1$ , just a removable discontinuity. Choice A incorrectly assumes an infinite limit without factoring. Factor completely and cancel before determining vertical asymptotes in rational functions.
If $q(x)=\dfrac{2}{x-4}+1$ , which statement correctly identifies the vertical asymptote using limits?
Answer: $\lim_{x\to4^-} q(x)=-\infty$ and $\lim_{x\to4^+} q(x)=+\infty$ , so $x=4$ is a vertical asymptote This problem tests linking infinite limits to vertical asymptotes in functions with added constants. For q(x) = 2/(x-4) + 1, the denominator zero at x=4 with nonzero numerator term drives the behavior. Approaching from the left yields negative infinity, and from the right positive infinity, despite the +1 not affecting the infinite trend. These opposing infinite limits confirm a vertical asymptote at x=4. A common distractor confuses this with a horizontal asymptote at y=1 due to the added constant, but horizontal asymptotes relate to limits at infinity, not vertical ones. A general strategy is to isolate terms causing division by zero and evaluate one-sided limits for infinity.
Let $p(x)=\dfrac{1}{(x+1)^2}$ . Which statement correctly identifies its vertical asymptote behavior?
Answer: $\lim_{x\to -1^-} p(x)=\infty$ and $\lim_{x\to -1^+} p(x)=\infty$ . This problem involves analyzing infinite limits for a rational function with a squared denominator. For $p(x)=\frac{1}{(x+1)^2}$ , the denominator equals zero when $x=-1$ , while the numerator remains 1 (non-zero). Since $(x+1)^2$ is always positive for $x\neq -1$ , as $x$ approaches $-1$ from either direction, the denominator approaches $0^+$ . This means $\lim_{x\to -1^-} p(x)=\frac{1}{0^+}=\infty$ and $\lim_{x\to -1^+} p(x)=\frac{1}{0^+}=\infty$ , confirming a vertical asymptote at $x=-1$ with the function approaching positive infinity from both sides. Choice C incorrectly suggests the limits have opposite signs, which would require the denominator to change sign. For functions with squared factors in the denominator, the vertical asymptote behavior is always the same from both sides.
For $f(x)=\dfrac{3x+1}{x^2-4}$ , which statement correctly describes the vertical asymptote behavior?
Answer: There are vertical asymptotes at $x=-2$ and $x=2$ . This question tests your ability to identify vertical asymptotes by analyzing when denominators equal zero and limits become infinite. For $f(x)=\frac{3x+1}{x^2-4}$ , we need to find where the denominator $x^2-4=0$ , which factors as $(x-2)(x+2)=0$ , giving us $x=2$ and $x=-2$ . At both these values, the numerator $3x+1$ is non-zero (it equals 7 and -5 respectively), so the function has infinite limits at these points. Since the denominator changes sign as we cross each zero, we get vertical asymptotes at both $x=-2$ and $x=2$ . Choice A incorrectly suggests an asymptote at $x=0$ where the function is actually well-defined. When finding vertical asymptotes, always check that the numerator doesn't also equal zero at the same point, as that could indicate a removable discontinuity instead.
For $r(x)=\dfrac{2}{x^2-4}$ , which statement correctly describes the vertical asymptote behavior near $x=2$ ?
Answer: $\lim_{x\to2^-}r(x)=-\infty$ and $\lim_{x\to2^+}r(x)=\infty$ This problem requires analyzing infinite limits for a rational function with a factorable denominator. For $r(x)=\frac{2}{x^2-4}=\frac{2}{(x-2)(x+2)}$ , near $x=2$ , the factor $(x+2)$ is approximately 4 (positive). When approaching from the left ( $x<2$ ), the factor $(x-2)$ is negative, making the denominator negative overall, so $r(x)\to-\infty$ . When approaching from the right ( $x>2$ ), the factor $(x-2)$ is positive, making the denominator positive, so $r(x)\to+\infty$ . This matches choice C, confirming that $x=2$ is a vertical asymptote with different one-sided infinite limits. Choice E incorrectly claims both limits are $+\infty$ , missing the sign change. The strategy is to determine the sign of each factor in the denominator on each side of the potential asymptote.
For $s(x)=\dfrac{x^2+1}{x^2-1}$ , which statement correctly identifies all vertical asymptotes?
Answer: There are vertical asymptotes at $x=-1$ and $x=1$ . This question tests your ability to find all vertical asymptotes by factoring and analyzing denominators. For $s(x)=\frac{x^2+1}{x^2-1}$ , we factor the denominator as $(x-1)(x+1)$ , which equals zero when $x=1$ or $x=-1$ . At both values, the numerator $x^2+1$ is non-zero (it equals 2 at both points). As $x\to 1$ , the factor $(x-1)$ changes sign, creating an infinite limit and vertical asymptote. Similarly, as $x\to -1$ , the factor $(x+1)$ changes sign, creating another vertical asymptote. Therefore, there are vertical asymptotes at both $x=-1$ and $x=1$ . Choice A incorrectly suggests only $x=0$ has an asymptote, but the function is well-defined there. To find all vertical asymptotes, factor the denominator completely and check each zero individually.
If $g(x)=\ln(x-5)$ , which limit indicates a vertical asymptote of $g$ ?
Answer: $\lim_{x\to 5^+} g(x)=-\infty$ . This problem requires understanding how logarithmic functions create vertical asymptotes at their domain boundaries through infinite limits. The function $g(x)=\ln(x-5)$ is only defined when $x-5>0$ , or $x>5$ , creating a domain boundary at $x=5$ . As $x$ approaches 5 from the right (the only valid direction), the argument $(x-5)$ approaches $0^+$ , and we know that $\ln(y)\to-\infty$ as $y\to 0^+$ . Therefore, $\lim_{x\to 5^+} g(x)=-\infty$ , indicating a vertical asymptote at $x=5$ . Choice E incorrectly suggests the limit from the left is positive infinity, but $g(x)$ isn't even defined for $x<5$ . For logarithmic functions, vertical asymptotes occur where the argument equals zero, and the function approaches negative infinity from the valid side.
For $q(x)=\dfrac{x^2-9}{x-3}$ , which statement correctly describes the behavior as $x\to3$ ?
Answer: There is a removable discontinuity at $x=3$ and $\lim_{x\to3} q(x)=6$ . This question tests your ability to connect infinite limits with vertical asymptotes via factoring and simplification. For q(x) = ( $x^2-9$ )/(x-3), factoring the numerator as (x-3)(x+3) allows cancellation with the denominator for x ≠ 3, simplifying to x+3. The limit as x approaches 3 is 6, which is finite, indicating no unbounded behavior. Thus, the original function has a removable discontinuity at x=3, not an asymptote. Choice A is tempting but fails because after simplification, there's no denominator zero, so the limit isn't infinite. Always factor and cancel common terms in rational functions to reveal whether the limit is finite or infinite.

Start this quiz