Calculus 1 Quiz: Lhospitals Rule
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Lhospitals RuleQuestion 1 of 9

Evaluate the limit lim⁡x→01x2∫0xtcos⁡(t) dt\lim_{x \to 0} \frac{1}{x^2} \int_0^x t \cos(t) \, dt.

00
13\frac{1}{3}
12\frac{1}{2}
The limit does not exist.
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Calculus 1 Quiz

Calculus 1 Quiz: Lhospitals Rule

Practice Lhospitals Rule in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Lhospitals Rule, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Question 1

Evaluate the limit lim⁡x→01x2∫0xtcos⁡(t) dt\lim_{x \to 0} \frac{1}{x^2} \int_0^x t \cos(t) \, dt.

  1. 00
  2. 13\frac{1}{3}
  3. 12\frac{1}{2} (correct answer)
  4. The limit does not exist.
Explanation: First, rewrite the limit as a quotient: lim⁡x→0∫0xtcos⁡(t) dtx2\lim_{x \to 0} \frac{\int_0^x t \cos(t) \, dt}{x^2} At x=0x=0, the numerator is ∫00tcos⁡(t) dt=0\int_0^0 t \cos(t) \, dt = 0 and the denominator is 02=00^2=0. The form is 00\frac{0}{0}. We apply L'Hospital's Rule. By the Fundamental Theorem of Calculus, Part 1, the derivative of the numerator is xcos⁡(x)x \cos(x). The derivative of the denominator is 2x2x. lim⁡x→0xcos⁡(x)2x=lim⁡x→0cos⁡(x)2\lim_{x \to 0} \frac{x \cos(x)}{2x} = \lim_{x \to 0} \frac{\cos(x)}{2} Now, we can evaluate by direct substitution: cos⁡(0)2=12.\frac{\cos(0)}{2} = \frac{1}{2}.

Question 2

What is the value of the limit lim⁡x→1+(x−1)tan⁡(πx2)\lim_{x \to 1^+} (x-1) \tan\left(\frac{\pi x}{2}\right)?

  1. −2π-\frac{2}{\pi} (correct answer)
  2. 2π\frac{2}{\pi}
  3. −π2-\frac{\pi}{2}
  4. 00
Explanation: The limit is of the indeterminate form 0⋅(−∞)0 \cdot (-\infty). To apply L'Hospital's Rule, we rewrite the expression as a quotient: lim⁡x→1+x−1cot⁡(πx2)\lim_{x \to 1^+} \frac{x-1}{\cot\left(\frac{\pi x}{2}\right)} This is now of the form 00\frac{0}{0}. Applying L'Hospital's Rule: lim⁡x→1+ddx(x−1)ddxcot⁡(πx2)=lim⁡x→1+1−csc⁡2(πx2)⋅π2\lim_{x \to 1^+} \frac{\frac{d}{dx}(x-1)}{\frac{d}{dx}\cot\left(\frac{\pi x}{2}\right)} = \lim_{x \to 1^+} \frac{1}{-\csc^2\left(\frac{\pi x}{2}\right) \cdot \frac{\pi}{2}} As x→1+x \to 1^+, πx2→π2\frac{\pi x}{2} \to \frac{\pi}{2}, and csc⁡(π2)=1\csc\left(\frac{\pi}{2}\right) = 1. So the limit is: 1−12⋅π2=−2π.\frac{1}{-1^2 \cdot \frac{\pi}{2}} = -\frac{2}{\pi}.

Question 3

Let ff be a twice-differentiable function. If lim⁡x→0f(x)−5xx2=3\lim_{x \to 0} \frac{f(x) - 5x}{x^2} = 3, and f(0)=0f(0)=0, what is the value of f′′(0)f''(0)?

  1. 33
  2. 55
  3. 66 (correct answer)
  4. Cannot be determined.
Explanation: The expression f(x)−5xx2\frac{f(x) - 5x}{x^2} has the indeterminate form 00\frac{0}{0} as x→0x \to 0 since f(0)=0f(0)=0. We can apply L'Hospital's Rule: lim⁡x→0f′(x)−52x=3\lim_{x \to 0} \frac{f'(x) - 5}{2x} = 3 Since the denominator approaches 0, for the limit to be a finite number (3), the numerator must also approach 0. Therefore, lim⁡x→0(f′(x)−5)=0\lim_{x \to 0} (f'(x) - 5) = 0, which implies f′(0)−5=0f'(0) - 5 = 0, so f′(0)=5f'(0) = 5. Now the limit expression lim⁡x→0f′(x)−52x\lim_{x \to 0} \frac{f'(x) - 5}{2x} is again of the indeterminate form 00\frac{0}{0}. We can apply L'Hospital's Rule a second time: lim⁡x→0f′′(x)2=3\lim_{x \to 0} \frac{f''(x)}{2} = 3 Since ff is twice-differentiable, f′′f'' is continuous at x=0x=0. Thus, we can substitute x=0x=0: f′′(0)2=3  ⟹  f′′(0)=6.\frac{f''(0)}{2} = 3 \implies f''(0) = 6.

Question 4

Let f(x)={2sin⁡(x)if x≥0a(ex−1)if x<0f(x) = \begin{cases} 2\sin(x) & \text{if } x \ge 0 \\ a(e^x - 1) & \text{if } x < 0 \end{cases}. Let g(x)=∫0xf(t) dtg(x) = \int_0^x f(t) \, dt. If f(x)f(x) is differentiable at x=0x=0, what is the value of lim⁡x→0g(x)x2\lim_{x \to 0} \frac{g(x)}{x^2}?

  1. 00
  2. 11 (correct answer)
  3. 22
  4. The limit depends on aa.
Explanation: For f(x)f(x) to be differentiable at x=0x=0, it must first be continuous. Continuity requires lim⁡x→0+f(x)=lim⁡x→0−f(x)\lim_{x\to 0^+} f(x) = \lim_{x\to 0^-} f(x), which means 2sin⁡(0)=a(e0−1)2\sin(0) = a(e^0-1), or 0=00=0. This is always true. For differentiability, the left and right derivatives must be equal at x=0x=0. f′(x)={2cos⁡(x)if x>0aexif x<0f'(x) = \begin{cases} 2\cos(x) & \text{if } x > 0 \\ ae^x & \text{if } x < 0 \end{cases}. At x=0x=0, we need 2cos⁡(0)=ae02\cos(0) = ae^0, which implies a=2a=2. So for f(x)f(x) to be differentiable, aa must be 2. Now we evaluate the limit lim⁡x→0g(x)x2\lim_{x \to 0} \frac{g(x)}{x^2}. This is of the form 00\frac{0}{0}. Using L'Hospital's Rule and the FTC: lim⁡x→0g′(x)2x=lim⁡x→0f(x)2x\lim_{x \to 0} \frac{g'(x)}{2x} = \lim_{x \to 0} \frac{f(x)}{2x} To evaluate this limit, we check from the left and right. Right-hand limit: lim⁡x→0+2sin⁡(x)2x=lim⁡x→0+sin⁡(x)x=1\lim_{x \to 0^+} \frac{2\sin(x)}{2x} = \lim_{x \to 0^+} \frac{\sin(x)}{x} = 1. Left-hand limit (with a=2a=2): lim⁡x→0−2(ex−1)2x=lim⁡x→0−ex−1x\lim_{x \to 0^-} \frac{2(e^x-1)}{2x} = \lim_{x \to 0^-} \frac{e^x-1}{x}. This is of the form 00\frac{0}{0}, so we use L'Hospital's Rule again: lim⁡x→0−ex1=1\lim_{x \to 0^-} \frac{e^x}{1} = 1. Since the left and right limits are both 1, the overall limit is 1.

Question 5

Evaluate lim⁡x→πsin⁡(x)x\lim_{x \to \pi} \frac{\sin(x)}{x}.

  1. 00 (correct answer)
  2. 11
  3. −1-1
  4. The limit does not exist.
Explanation: This limit is not an indeterminate form. We can evaluate it by direct substitution: lim⁡x→πsin⁡(x)x=sin⁡(π)π=0π=0.\lim_{x \to \pi} \frac{\sin(x)}{x} = \frac{\sin(\pi)}{\pi} = \frac{0}{\pi} = 0. L'Hospital's Rule does not apply because the form is 0π\frac{0}{\pi}, not 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}. A common mistake is to mechanically apply L'Hospital's Rule, which would lead to the incorrect result lim⁡x→πcos⁡(x)1=cos⁡(π)=−1\lim_{x \to \pi} \frac{\cos(x)}{1} = \cos(\pi) = -1.

Question 6

Let ff be a twice-differentiable function such that f(1)=0f(1)=0, f′(1)=0f'(1)=0, and f′′(1)=6f''(1)=6. Evaluate lim⁡x→1f(x)(x−1)2\lim_{x \to 1} \frac{f(x)}{(x-1)^2}.

  1. 00
  2. 33 (correct answer)
  3. 66
  4. 1212
Explanation: The limit is of the form f(1)(1−1)2=00\frac{f(1)}{(1-1)^2} = \frac{0}{0}. We apply L'Hospital's Rule: lim⁡x→1f′(x)2(x−1)\lim_{x \to 1} \frac{f'(x)}{2(x-1)} Since f′(1)=0f'(1)=0, this is also of the form 00\frac{0}{0}. We apply L'Hospital's Rule a second time: lim⁡x→1f′′(x)2\lim_{x \to 1} \frac{f''(x)}{2} Since ff is twice-differentiable, f′′f'' is continuous near 1, so we can substitute x=1x=1: f′′(1)2=62=3.\frac{f''(1)}{2} = \frac{6}{2} = 3.

Question 7

Evaluate lim⁡x→1(1ln⁡x−1x−1)\lim_{x \to 1} \left( \frac{1}{\ln x} - \frac{1}{x-1} \right).

  1. 00
  2. 12\frac{1}{2} (correct answer)
  3. 11
  4. The limit does not exist.
Explanation: The limit is of the indeterminate form ∞−∞\infty - \infty. First, combine the fractions: lim⁡x→1x−1−ln⁡x(ln⁡x)(x−1)\lim_{x \to 1} \frac{x-1 - \ln x}{(\ln x)(x-1)} As x→1x \to 1, this becomes the indeterminate form 00\frac{0}{0}. Applying L'Hospital's Rule: lim⁡x→1ddx(x−1−ln⁡x)ddx((ln⁡x)(x−1))=lim⁡x→11−1x1x(x−1)+(ln⁡x)(1)=lim⁡x→1x−1xx−1x+ln⁡x\lim_{x \to 1} \frac{\frac{d}{dx}(x-1 - \ln x)}{\frac{d}{dx}((\ln x)(x-1))} = \lim_{x \to 1} \frac{1 - \frac{1}{x}}{\frac{1}{x}(x-1) + (\ln x)(1)} = \lim_{x \to 1} \frac{\frac{x-1}{x}}{\frac{x-1}{x} + \ln x} Multiplying the numerator and denominator by xx simplifies this to: lim⁡x→1x−1x−1+xln⁡x\lim_{x \to 1} \frac{x-1}{x-1 + x \ln x} This is still of the form 00\frac{0}{0}. Applying L'Hospital's Rule again: lim⁡x→111+(1⋅ln⁡x+x⋅1x)=lim⁡x→111+ln⁡x+1=lim⁡x→112+ln⁡x=12+0=12.\lim_{x \to 1} \frac{1}{1 + (1 \cdot \ln x + x \cdot \frac{1}{x})} = \lim_{x \to 1} \frac{1}{1 + \ln x + 1} = \lim_{x \to 1} \frac{1}{2 + \ln x} = \frac{1}{2+0} = \frac{1}{2}.

Question 8

Evaluate lim⁡x→0+(ex+2x)1/x\lim_{x \to 0^+} (e^x + 2x)^{1/x}.

  1. e3e^3 (correct answer)
  2. e2e^2
  3. 33
  4. 11
Explanation: This limit is of the indeterminate form 1∞1^\infty. Let L=lim⁡x→0+(ex+2x)1/xL = \lim_{x \to 0^+} (e^x + 2x)^{1/x}. We take the natural logarithm of both sides: ln⁡(L)=lim⁡x→0+1xln⁡(ex+2x)=lim⁡x→0+ln⁡(ex+2x)x\ln(L) = \lim_{x \to 0^+} \frac{1}{x} \ln(e^x + 2x) = \lim_{x \to 0^+} \frac{\ln(e^x + 2x)}{x} This is of the form 00\frac{0}{0}. Applying L'Hospital's Rule: ln⁡(L)=lim⁡x→0+ex+2ex+2x1=e0+2e0+0=1+21=3.\ln(L) = \lim_{x \to 0^+} \frac{\frac{e^x+2}{e^x+2x}}{1} = \frac{e^0+2}{e^0+0} = \frac{1+2}{1} = 3. Since ln⁡(L)=3\ln(L) = 3, the original limit is L=e3L = e^3.

Question 9

Evaluate the limit: L=lim⁡x→0e2x−e−2x−4xx−sin⁡(x)L = \lim_{x \to 0} \frac{e^{2x} - e^{-2x} - 4x}{x - \sin(x)}

  1. L=0L = 0
  2. L=4L = 4
  3. L=8L = 8
  4. L=16L = 16 (correct answer)
Explanation: The limit is of the indeterminate form 00\frac{0}{0}. Applying L'Hospital's Rule: lim⁡x→02e2x+2e−2x−41−cos⁡(x)\lim_{x \to 0} \frac{2e^{2x} + 2e^{-2x} - 4}{1 - \cos(x)} This is still of the form 00\frac{0}{0}. Applying L'Hospital's Rule a second time: lim⁡x→04e2x−4e−2xsin⁡(x)\lim_{x \to 0} \frac{4e^{2x} - 4e^{-2x}}{\sin(x)} This is still of the form 00\frac{0}{0}. Applying L'Hospital's Rule a third time: lim⁡x→08e2x+8e−2xcos⁡(x)=8e0+8e0cos⁡(0)=8+81=16.\lim_{x \to 0} \frac{8e^{2x} + 8e^{-2x}}{\cos(x)} = \frac{8e^0 + 8e^0}{\cos(0)} = \frac{8+8}{1} = 16.