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.

This question tests your ability to connect infinite limits with vertical asymptotes when one-sided limits differ. If the left-hand limit at x=-3 is negative infinity and the right-hand is finite (5), the infinite side indicates unbounded behavior approaching x=-3. This supports a vertical asymptote, even if only one side is infinite. The finite side means the graph approaches 5 from the right, but the infinite side dominates for asymptote classification. Choice E is tempting but fails because vertical asymptotes require at least one infinite one-sided limit, not necessarily both. Always evaluate both one-sided limits separately; an infinite result from either confirms a vertical asymptote.

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.

This question tests recognizing when a rational function has a removable discontinuity rather than a vertical asymptote. For $h(x)=\frac{x^2-9}{x-3}$, we can factor the numerator as $(x-3)(x+3)$, giving $h(x)=\frac{(x-3)(x+3)}{x-3}$. When $x\neq 3$, this simplifies to $h(x)=x+3$, so $\lim_{x\to 3}h(x)=3+3=6$. Since the limit exists and is finite, there is no vertical asymptote at $x=3$; instead, this is a removable discontinuity (a hole in the graph). Choice A incorrectly assumes that whenever a denominator equals zero, there must be a vertical asymptote with an infinite limit. The key insight is that when factors cancel between numerator and denominator, the result is a removable discontinuity with a finite limit, not a vertical asymptote.

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.

This problem requires recognizing how factor cancellation affects vertical asymptotes through infinite limit analysis. The function g(x) = [(x-4)(x+1)]/[(x-4)(x-2)] has a common factor of (x-4) in both numerator and denominator that cancels, leaving g(x) = (x+1)/(x-2) for x ≠ 4. After cancellation, the only zero of the denominator is x = 2, where the simplified function approaches ±∞. Therefore, there is a vertical asymptote only at x = 2, not at x = 4. Choice A incorrectly assumes vertical asymptotes at both x = 4 and x = 2, failing to recognize that the cancellation at x = 4 creates a removable discontinuity instead. Remember that vertical asymptotes occur only where the denominator approaches zero without a corresponding zero in the numerator after all possible cancellations.

This question examines vertical asymptotes in logarithmic functions through infinite limit behavior. The function h(x) = ln(x-3) is defined only for x > 3, where the argument (x-3) must be positive. As x approaches 3 from the right (x → 3⁺), the argument (x-3) approaches 0 through positive values, and ln(x-3) approaches -∞. This creates a vertical asymptote at x = 3 with the specific one-sided limit behavior described in choice A. Choice B incorrectly places the asymptote at y = 3 instead of x = 3, confusing vertical and horizontal asymptotes. For logarithmic functions ln(x-a), vertical asymptotes always occur at x = a where the argument becomes zero.

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.

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.

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.