Selecting Limit Procedures - Calculus 1

1. Consider $\lim_{x \to \infty} \frac{3^x}{x^3}$. Which procedure is appropriate?
   Answer: Apply L'Hôpital's Rule three times. When you encounter limits involving exponential and polynomial functions, you're dealing with a classic indeterminate form that requires careful analysis of growth rates. This limit has the form $\frac{\infty}{\infty}$ since both $3^x$ and $x^3$ approach infinity as $x \to \infty$. This indeterminate form is perfect for L'Hôpital's Rule, which states that if $\lim_{x \to a} \frac{f(x)}{g(x)}$ gives $\frac{\infty}{\infty}$, then the limit equals $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ (if this limit exists). Applying L'Hôpital's Rule: First application gives $\frac{3^x \ln 3}{3x^2}$, still $\frac{\infty}{\infty}$. Second application: $\frac{3^x (\ln 3)^2}{6x}$, still indeterminate. Third application: $\frac{3^x (\ln 3)^3}{6} = \infty$. Answer D is correct. Option A fails because dividing by $x^3$ gives $\frac{(3/x)^x \cdot 3^x}{x^3} \cdot \frac{1/x^3}{1/x^3}$, which doesn't simplify the exponential growth issue. Option B incorrectly suggests taking the logarithm of the entire expression, but we need to evaluate the limit of the original expression, not its logarithm. Option C shows a fundamental misunderstanding—when both numerator and denominator approach infinity, you can't conclude the limit is 1; exponential functions grow much faster than polynomials. Remember: exponential functions always dominate polynomial functions as $x \to \infty$. When you see exponential over polynomial, expect L'Hôpital's Rule and prepare to apply it multiple times.
2. To evaluate $\lim_{x \to 0^+} (\cot x)^{\sin x}$, which sequence of procedures is most appropriate?
   Answer: Let $y = (\cot x)^{\sin x}$, take the natural logarithm, and transform the resulting product into a quotient. The limit has the indeterminate form $\infty^0$. To resolve this, we must use logarithms. Let $y = (\cot x)^{\sin x}$. Then $\ln y = \sin x \ln(\cot x)$. This is a $0 \cdot \infty$ form, which must be rewritten as a quotient (e.g., $\frac{\ln(\cot x)}{\csc x}$) before L'Hôpital's Rule can be applied.
3. For the function $f(x) = \frac{|x-3|}{x^2-9}$, which procedure is required to determine $\lim_{x \to 3} f(x)$?
   Answer: Evaluate the left-hand and right-hand limits separately and compare them. When evaluating limits involving absolute value functions, you need to consider that absolute value expressions behave differently on either side of their critical point. Since the absolute value changes its definition at $x = 3$, you must examine the limit from both directions. For $x < 3$, we have $|x-3| = -(x-3) = 3-x$, so the function becomes: $f(x) = \frac{3-x}{x^2-9} = \frac{3-x}{(x-3)(x+3)} = \frac{-(x-3)}{(x-3)(x+3)} = \frac{-1}{x+3}$ For $x > 3$, we have $|x-3| = x-3$, so: $f(x) = \frac{x-3}{x^2-9} = \frac{x-3}{(x-3)(x+3)} = \frac{1}{x+3}$ The left-hand limit is $\lim_{x \to 3^-} \frac{-1}{x+3} = \frac{-1}{6}$, while the right-hand limit is $\lim_{x \to 3^+} \frac{1}{x+3} = \frac{1}{6}$. Since these are unequal, the limit doesn't exist. Option A is wrong because while substitution gives $\frac{0}{0}$, L'Hôpital's Rule doesn't apply directly to absolute value functions without first analyzing the cases. Option B fails because you can't simply cancel $x-3$ when the numerator contains $|x-3|$, which isn't always equal to $x-3$. Option C is incorrect since the conjugate method applies to expressions with square roots, not absolute values. Study tip: Whenever you see absolute value in a limit problem, immediately check if the limit point is where the absolute value expression equals zero—this signals you need one-sided limits.
4. Consider the limit $L = \lim_{x\to\infty} \frac{x + \sin(x)}{2x + \cos(x)}$. A student attempts to use L'Hôpital's Rule because the limit appears to be of the indeterminate form $\frac{\infty}{\infty}$. Which statement best explains why this approach is problematic and identifies a more reliable procedure?
   Answer: Applying L'Hôpital's Rule leads to a limit that does not exist, so the rule is inconclusive; dividing the numerator and denominator by $x$ is the appropriate method. Direct substitution leads to the indeterminate form $\frac{\infty}{\infty}$, so L'Hôpital's Rule seems applicable. Applying the rule gives $\lim_{x\to\infty} \frac{1 + \cos(x)}{2 - \sin(x)}$. This new limit does not exist because the numerator oscillates between 0 and 2, and the denominator oscillates between 1 and 3. When the limit of the ratio of derivatives does not exist, L'Hôpital's Rule is inconclusive. A more reliable procedure is to divide the numerator and denominator by the highest power of $x$: $\lim_{x\to\infty} \frac{1 + \frac{\sin(x)}{x}}{2 + \frac{\cos(x)}{x}}$. Since $\lim_{x\to\infty} \frac{\sin(x)}{x} = 0$ and $\lim_{x\to\infty} \frac{\cos(x)}{x} = 0$, the limit is $\frac{1+0}{2+0} = \frac{1}{2}$.
5. What is the most direct procedure to evaluate $\lim_{x \to \pi/2} \frac{\cot(x)}{\csc(x)}$?
   Answer: Rewrite the expression in terms of sine and cosine and simplify algebraically. The most efficient approach is algebraic simplification. $\frac{\cot(x)}{\csc(x)} = \frac{\cos(x)/\sin(x)}{1/\sin(x)} = \cos(x)$. Therefore, $\lim_{x \to \pi/2} \frac{\cot(x)}{\csc(x)} = \lim_{x \to \pi/2} \cos(x) = 0$. This avoids dealing with the indeterminate forms that would arise from working with cotangent and cosecant directly.
6. The limit $L = \lim_{h\to 0} \frac{\tan(\pi/4 + h) - 1}{h}$ can be evaluated using several methods. Which procedure provides the most direct evaluation by recognizing the limit's fundamental structure?
   Answer: Identifying the limit as the definition of the derivative for a specific function at a specific point. The expression is in the form $\lim_{h\to 0} \frac{f(a+h) - f(a)}{h}$, which is the definition of the derivative $f'(a)$. Here, $f(x) = \tan(x)$ and $a = \pi/4$. Note that $f(\pi/4) = \tan(\pi/4) = 1$. Therefore, the limit is equivalent to finding the derivative of $\tan(x)$ at $x=\pi/4$. Since $f'(x) = \sec^2(x)$, the limit is $\sec^2(\pi/4) = (\sqrt{2})^2 = 2$. While L'Hôpital's Rule (A) and using trig identities (C) are also valid procedures that yield the correct answer, recognizing the definition of the derivative is the most direct conceptual approach.
7. What is the most direct and efficient procedure for determining the limit $\lim_{x \to \infty} \frac{8x^4 - 3x^2}{ \sqrt{16x^8 + x^3}}$ ?
   Answer: Divide both the numerator and the denominator by $x^4$. For limits of rational-like functions at infinity, the most efficient method is to divide the numerator and denominator by the highest power of $x$ in the denominator. Here, the effective power in the denominator is $\sqrt{x^8} = x^4$. Dividing all terms by $x^4$ simplifies the expression quickly. While L'Hôpital's Rule is applicable (form is $\infty/\infty$), it would be extremely inefficient.
8. Consider the limit $\lim_{x \to a} \frac{f(x)}{g(x)}$. It is known that $\lim_{x \to a} f(x) = K \neq 0$ and $\lim_{x \to a} g(x) = 0$. Which procedure must be followed?
   Answer: Investigate the one-sided limits by analyzing the sign of $g(x)$ near $a$. The limit is of the form $K/0$ with $K \neq 0$. This indicates the presence of a vertical asymptote. The limit could be $+\infty$, $-\infty$, or it may not exist if the one-sided limits differ. To determine the behavior, one must evaluate the left-hand and right-hand limits by checking the sign of the denominator as $x$ approaches $a$ from each side.
9. Let $p(x)$ be a polynomial of degree $n \ge 1$. To evaluate $\lim_{x \to \infty} \frac{p(x)}{e^{2x}}$ using L'Hôpital's Rule, which statement correctly describes the process?
   Answer: L'Hôpital's Rule must be applied exactly $n$ times. The limit is of the form $\infty/\infty$. Each application of L'Hôpital's Rule reduces the degree of the polynomial numerator by one. This process must be repeated $n$ times until the numerator becomes a constant. At that point, the limit will be $\lim_{x \to \infty} \frac{\text{constant}}{k \cdot e^{2x}}$, which is 0.
10. To evaluate the limit $\lim_{x \to \infty} (\sqrt{4x^2 + 5x} - 2x)$, which of the following is the most appropriate initial step?
    Answer: Multiply the expression by its conjugate, $\frac{\sqrt{4x^2 + 5x} + 2x}{\sqrt{4x^2 + 5x} + 2x}$. The limit presents the indeterminate form $\infty - \infty$. The standard and most effective procedure for limits of this type involving square roots is to multiply by the conjugate. This will transform the expression into a rational function, which can then be evaluated by dividing by the highest power of $x$.
11. For $\lim_{x \to 0} (\cos x)^{1/x^2}$, what is the correct initial procedure?
    Answer: Let $y = (\cos x)^{1/x^2}$ and evaluate the limit of $\ln y$. When you encounter a limit involving an expression raised to a power, especially when direct substitution gives an indeterminate form, you're dealing with a situation that requires the logarithmic technique. Here, substituting $x = 0$ gives $(\cos 0)^{1/0^2} = 1^{\infty}$, which is indeterminate. The correct approach is option D: let $y = (\cos x)^{1/x^2}$ and evaluate $\lim_{x \to 0} \ln y$. Taking the natural logarithm gives $\ln y = \frac{\ln(\cos x)}{x^2}$. As $x \to 0$, this becomes $\frac{0}{0}$, which is perfect for L'Hôpital's Rule. Once you find $\lim_{x \to 0} \ln y = L$, then $\lim_{x \to 0} y = e^L$. Option A is incorrect because you cannot apply L'Hôpital's Rule to the base and exponent separately when dealing with indeterminate forms like $1^{\infty}$. The rule applies to quotients, not to bases and exponents independently. Option B uses an irrelevant trigonometric identity that doesn't simplify the indeterminate form. The identity $\cos x = \sqrt{1-\sin^2 x}$ would actually complicate the expression rather than resolve the $1^{\infty}$ issue. Option C incorrectly assumes that $1^{\infty} = 1$, but this is an indeterminate form, not a definite value. The limit could be any positive real number or infinity, depending on how the base approaches 1 and the exponent approaches infinity. Key takeaway: For limits of the form $f(x)^{g(x)}$ that yield indeterminate forms, always use the logarithmic technique first.
12. Which sequence of procedures is most appropriate for evaluating $\lim_{x\to 0} \frac{e^x - 1 - \sin(x)}{1 - \cos(x)}$?
    Answer: Apply L'Hôpital's Rule twice, as the indeterminate form $\frac{0}{0}$ persists after the first application. When evaluating limits that result in indeterminate forms like $\frac{0}{0}$, you need to systematically work through the algebra or apply techniques like L'Hôpital's Rule until you reach a determinate form. Let's verify this is indeed $\frac{0}{0}$: as $x \to 0$, both $e^x - 1 - \sin(x) \to 0$ and $1 - \cos(x) \to 0$. Applying L'Hôpital's Rule once gives us: $\lim_{x\to 0} \frac{e^x - \cos(x)}{\sin(x)}$ Checking again: as $x \to 0$, the numerator approaches $1 - 1 = 0$ and the denominator approaches $0$, so we still have $\frac{0}{0}$. Applying L'Hôpital's Rule a second time: $\lim_{x\to 0} \frac{e^x + \sin(x)}{\cos(x)} = \frac{1 + 0}{1} = 1$ Therefore, A is correct - you need L'Hôpital's Rule twice because the indeterminate form persists after the first application. B is incorrect because multiplying by the conjugate $(1 + \cos(x))$ would create $\sin^2(x)$ in the denominator, which doesn't simplify the original expression effectively. C fails because after one application of L'Hôpital's Rule, you get $\frac{e^x - \cos(x)}{\sin(x)}$, which is still $\frac{0}{0}$ and cannot be resolved using just the standard sine limit. D is problematic because when you separate the fractions, each individual limit $\frac{e^x - 1}{1 - \cos(x)}$ and $\frac{\sin(x)}{1 - \cos(x)}$ doesn't exist in elementary form, making this approach unnecessarily complex. Study tip: When L'Hôpital's Rule gives you another indeterminate form, don't hesitate to apply it again - some limits require multiple applications to resolve.
13. To evaluate $\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{x}$, which two procedures are both valid initial steps?
    Answer: Applying L'Hôpital's Rule and multiplying by the conjugate. When you encounter a limit that produces an indeterminate form like $\frac{0}{0}$, you have several tools at your disposal. Let's check what happens with direct substitution: $\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{x} = \frac{\sqrt{1} - \sqrt{1}}{0} = \frac{0}{0}$. This indeterminate form means you need alternative approaches. Two methods work perfectly here. L'Hôpital's Rule applies since we have $\frac{0}{0}$. Taking derivatives: $\lim_{x \to 0} \frac{\frac{1}{2\sqrt{1+x}} - \frac{-1}{2\sqrt{1-x}}}{1} = \frac{1}{2} + \frac{1}{2} = 1$. Multiplying by the conjugate also works beautifully with square root differences: $\frac{\sqrt{1+x} - \sqrt{1-x}}{x} \cdot \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} = \frac{(1+x)-(1-x)}{x(\sqrt{1+x} + \sqrt{1-x})} = \frac{2x}{x(\sqrt{1+x} + \sqrt{1-x})} = \frac{2}{\sqrt{1+x} + \sqrt{1-x}}$, which gives $\frac{2}{2} = 1$. Choice A incorrectly suggests the Squeeze Theorem, which requires finding bounding functions—unnecessary and overly complicated here. Choice B mentions factoring the denominator, but $x$ is already in simplest form, making this impossible. Choice C incorrectly claims direct substitution works as an initial step, but we showed it gives $\frac{0}{0}$. Choice D correctly identifies both L'Hôpital's Rule and the conjugate method as valid approaches. Study tip: For limits involving square root differences, always consider the conjugate method first—it often eliminates the indeterminate form elegantly. Keep L'Hôpital's Rule as your backup for $\frac{0}{0}$ situations.
14. Let $f$ and $g$ be differentiable functions such that $\lim_{x\to a} f(x) = \infty$, $\lim_{x\to a} g(x) = \infty$, and $\lim_{x\to a} \frac{f'(x)}{g'(x)} = L$. What procedure is justified by these conditions to find $\lim_{x\to a} \frac{f(x)}{g(x)}$?
    Answer: L'Hôpital's Rule, because the limit is of the indeterminate form $\frac{\infty}{\infty}$ and the limit of the ratio of derivatives exists. The given conditions are the exact premises required to apply L'Hôpital's Rule for the indeterminate form $\frac{\infty}{\infty}$. The rule states that if $\lim_{x\to a} f(x) = \pm \infty$ and $\lim_{x\to a} g(x) = \pm \infty$, and if $\lim_{x\to a} \frac{f'(x)}{g'(x)}$ exists (and equals $L$), then $\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)} = L$. Therefore, L'Hôpital's Rule is the justified procedure.
15. Which statement provides a valid initial procedure and justification for evaluating $\lim_{x \to 3} \frac{x^3 - 27}{x^2 - 9}$?
    Answer: Because direct substitution fails, factor both polynomials and cancel the common term. Direct substitution gives $0/0$, an indeterminate form. For rational functions, this indicates a common factor in the numerator and denominator. Factoring gives $\lim_{x \to 3} \frac{(x-3)(x^2+3x+9)}{(x-3)(x+3)}$. Canceling the $(x-3)$ term is a valid and effective procedure. Applying L'Hôpital's Rule is also valid, but factoring is often considered more fundamental for polynomials.
16. Which procedure must be used to determine $\lim_{x \to 0} f(x)$?
    Answer: Evaluate the limits of the bounding functions and apply the Squeeze Theorem. The problem provides upper and lower bounds for $f(x)$. The appropriate procedure for this situation is the Squeeze Theorem. We evaluate the limits of the bounds: $\lim_{x \to 0} (e^x - 1) = 1 - 1 = 0$ and $\lim_{x \to 0} (x^2+x) = 0$. Since both limits are equal, the Squeeze Theorem guarantees that $\lim_{x \to 0} f(x) = 0$.
17. Given that $\lim_{x \to 0} \frac{1-\cos x}{x^2} = \frac{1}{2}$, what procedure justifies the conclusion that $\lim_{x \to 0} \frac{x^2}{1-\cos x} = 2$ without a full re-evaluation?
    Answer: Using the limit property $\lim [1/g(x)] = 1 / [\lim g(x)]$ for a non-zero limit. The second expression, $\frac{x^2}{1-\cos x}$, is the reciprocal of the first expression. Since the limit of the first expression is $1/2$ (which is not zero), the limit of its reciprocal is the reciprocal of the limit. Therefore, $\lim_{x \to 0} \frac{x^2}{1-\cos x} = \frac{1}{\lim_{x \to 0} \frac{1-\cos x}{x^2}} = \frac{1}{1/2} = 2$.
18. A student wants to evaluate $\lim_{x \to 0^+} (x^2 \ln x)$ using L'Hôpital's Rule. Which of the following transformations correctly sets up the problem for applying this rule?
    Answer: Rewrite the expression as $\lim_{x \to 0^+} \frac{\ln x}{1/x^2}$. The original limit is of the indeterminate form $0 \cdot (-\infty)$. To use L'Hôpital's Rule, it must be converted to a quotient of the form $0/0$ or $\infty/\infty$. Rewriting as $\frac{\ln x}{1/x^2}$ creates the form $-\infty/\infty$, which is suitable for L'Hôpital's Rule. Option B is also a valid transformation but leads to a more complicated derivative.
19. A student attempts to find $\lim_{x \to 1} \frac{x^2+3}{x-1}$ by applying L'Hôpital's Rule, calculating $\lim_{x \to 1} \frac{2x}{1} = 2$. What is the fundamental error in this procedure?
    Answer: L'Hôpital's Rule was applied to a limit not of the form $0/0$ or $\infty/\infty$. L'Hôpital's Rule is only applicable when direct substitution results in an indeterminate form of $0/0$ or $\infty/\infty$. In this case, direct substitution yields $4/0$. This is not an indeterminate form, so L'Hôpital's Rule cannot be used. The correct procedure is to analyze the one-sided limits, which approach $\pm\infty$.
20. To evaluate $\lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{2x}$, which of the following describes the necessary first step?
    Answer: Set $y$ equal to the expression and evaluate the limit of $\ln(y)$ to handle the indeterminate form. The limit is of the indeterminate form $1^\infty$. This form must be manipulated before a limit can be found. The standard procedure is to let $y = (1 + 3/x)^{2x}$, take the natural logarithm of both sides ($\ln y = 2x \ln(1+3/x)$), and then evaluate $\lim_{x \to \infty} \ln y$. This transforms the problem into a $\infty \cdot 0$ form, which can be rearranged for L'Hôpital's Rule.

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