Limits and Derivatives MCQs in Calculus

By: Prof. Dr. Fazal Rehman Shamil | Last updated: February 12, 2025

\[
\textbf{Question 1:}
\]
\[
\text{What is the limit of } \lim_{x \to 0} \frac{\sin(x)}{x}?
\]
\[
\text{(a) } 0, \quad \text{(b) } 1, \quad \text{(c) } \infty, \quad \text{(d) } -1
\]
\[
\text{Answer: B}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]
\[
\text{The standard limit } \lim_{x \to 0} \frac{\sin(x)}{x} \text{ is a fundamental result in calculus.}
\]
\[
\text{Using L’Hôpital’s Rule:}
\]
\[
\lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \frac{\cos(x)}{1} = 1.
\]
\[
\text{Thus, the correct answer is } \boxed{1}.
\]

\[
\textbf{Question 2:}
\]
\[
\text{Find the derivative of } f(x) = 4x^3 – 5x^2 + 2x – 1.
\]
\[
\text{(a) } 12x^2 – 10x + 2, \quad \text{(b) } 12x^2 – 10x, \quad \text{(c) } 12x^2 + 10x, \quad \text{(d) } 4x^3 – 5x
\]
\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]
\[
\text{Differentiate term by term:}
\]
\[
\frac{d}{dx} (4x^3 – 5x^2 + 2x – 1) = 12x^2 – 10x + 2.
\]
\[
\text{Thus, the correct answer is } \boxed{12x^2 – 10x + 2}.
\]

\[
\textbf{Question 3:}
\]
\[
\text{What is the limit of } \lim_{x \to 2} \frac{x^2 – 4}{x – 2}?
\]
\[
\text{(a) } 0, \quad \text{(b) } 2, \quad \text{(c) } 4, \quad \text{(d) } \infty
\]
\[
\text{Answer: B}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]
\[
\text{Factor the numerator:}
\]
\[
\frac{x^2 – 4}{x – 2} = \frac{(x – 2)(x + 2)}{x – 2}.
\]
\[
\text{Cancel } (x – 2):
\]
\[
= x + 2.
\]
\[
\text{Substituting } x = 2, \text{ we get } 2 + 2 = 4.
\]
\[
\text{Thus, the correct answer is } \boxed{4}.
\]

\[
\textbf{Question 4:}
\]
\[
\text{Which of the following represents the derivative of } f(x) = e^x?
\]
\[
\text{(a) } e^x, \quad \text{(b) } 1, \quad \text{(c) } x, \quad \text{(d) } \ln(x)
\]
\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]
\[
\text{The derivative of } e^x \text{ is } e^x.
\]
\[
\frac{d}{dx} e^x = e^x.
\]
\[
\text{Thus, the correct answer is } \boxed{e^x}.
\]

\[
\textbf{Question 5:}
\]
\[
\text{Find the derivative of } f(x) = \ln(x).
\]
\[
\text{(a) } \frac{1}{x}, \quad \text{(b) } x, \quad \text{(c) } \frac{1}{x^2}, \quad \text{(d) } x \ln(x)
\]
\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]
\[
\text{The derivative of } \ln(x) \text{ is given by:}
\]
\[
\frac{d}{dx} \ln(x) = \frac{1}{x}.
\]
\[
\text{Thus, the correct answer is } \boxed{\frac{1}{x}}.
\]

\[
\textbf{Question 6:}
\]
\[
\text{What is the derivative of } f(x) = \cos(x)?
\]
\[
\text{(a) } \sin(x), \quad \text{(b) } -\sin(x), \quad \text{(c) } \cos(x), \quad \text{(d) } -\cos(x)
\]
\[
\text{Answer: B}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]
\[
\text{The derivative of } \cos(x) \text{ is } -\sin(x).
\]
\[
\frac{d}{dx} \cos(x) = -\sin(x).
\]
\[
\text{Thus, the correct answer is } \boxed{-\sin(x)}.
\]

\[
\textbf{Question 7:}
\]
\[
\text{Find the limit of } \lim_{x \to 0} \frac{e^x – 1}{x}.
\]
\[
\text{(a) } 1, \quad \text{(b) } 0, \quad \text{(c) } \infty, \quad \text{(d) } -1
\]
\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]
\[
\text{Using L’Hôpital’s Rule:}
\]
\[
\lim_{x \to 0} \frac{e^x – 1}{x} = \lim_{x \to 0} \frac{e^x}{1} = e^0 = 1.
\]
\[
\text{Thus, the correct answer is } \boxed{1}.
\]

\[
\textbf{Question 8:}
\]
\[
\text{Which rule is used to find the derivative of a product of two functions?}
\]
\[
\text{(a) } \text{Chain Rule}, \quad \text{(b) } \text{Product Rule}, \quad \text{(c) } \text{Quotient Rule}, \quad \text{(d) } \text{Power Rule}
\]
\[
\text{Answer: B}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]
\[
\text{The Product Rule states: } \frac{d}{dx} [uv] = u’v + uv’.
\]
\[
\text{Thus, the correct answer is } \boxed{\text{Product Rule}}.
\]

\[
\textbf{Question 9:}
\]
\[
\text{What is the limit of } \lim_{x \to 0} \frac{1 – \cos(x)}{x^2}?
\]
\[
\text{(a) } 0, \quad \text{(b) } 1, \quad \text{(c) } \infty, \quad \text{(d) } \frac{1}{2}
\]
\[
\text{Answer: D}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]
\[
\text{Using the Taylor series expansion: } \cos(x) \approx 1 – \frac{x^2}{2}.
\]
\[
\frac{1 – \cos(x)}{x^2} = \frac{x^2 / 2}{x^2} = \frac{1}{2}.
\]
\[
\text{Thus, the correct answer is } \boxed{\frac{1}{2}}.
\]

\[
\textbf{Question 10:}
\]
\[
\text{Find the derivative of } f(x) = x^5 + 3x^4 – x^3 + 2.
\]
\[
\text{(a) } 5x^4 + 12x^3 – 3x^2, \quad \text{(b) } 5x^4 + 12x^3 – x^2, \quad \text{(c) } 5x^5 + 12x^4 – x^3, \quad \text{(d) } 5x^4 + 3x^3 – 2x
\]
\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]
\[
\text{Differentiate term by term:}
\]
\[
\frac{d}{dx} (x^5 + 3x^4 – x^3 + 2) = 5x^4 + 12x^3 – 3x^2.
\]
\[
\text{Thus, the correct answer is } \boxed{5x^4 + 12x^3 – 3x^2}.
\]

\[
\textbf{Question 11:}
\]
\[
\text{What is the derivative of } f(x) = \tan(x)?
\]
\[
\text{(a) } \sec^2(x), \quad \text{(b) } \cos^2(x), \quad \text{(c) } \sin^2(x), \quad \text{(d) } \sec(x)
\]
\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]
\[
\text{The derivative of } \tan(x) \text{ is } \sec^2(x).
\]
\[
\frac{d}{dx} \tan(x) = \sec^2(x).
\]
\[
\text{Thus, the correct answer is } \boxed{\sec^2(x)}.
\]

\[
\textbf{Question 12:}
\]
\[
\text{Find the limit of } \lim_{x \to \infty} \frac{1}{x^2}.
\]
\[
\text{(a) } 0, \quad \text{(b) } 1, \quad \text{(c) } \infty, \quad \text{(d) } -1
\]
\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]
\[
\text{As } x \to \infty, \text{ the denominator } x^2 \text{ grows very large.}
\]
\[
\text{Thus, } \frac{1}{x^2} \to 0.
\]
\[
\text{Thus, the correct answer is } \boxed{0}.
\]

\[
\textbf{Question 13:}
\]
\[
\text{Find the derivative of } f(x) = 5x^3 + 4x^2 – 6x + 7.
\]
\[
\text{(a) } 15x^2 + 8x – 6, \quad \text{(b) } 15x^2 + 4x – 6, \quad \text{(c) } 5x^3 + 4x^2 + 6, \quad \text{(d) } 5x^3 + 8x^2 – 6
\]
\[
\text{Answer: A}
\]

Step by Step Solution

\[
\textbf{Solution:}
\]
\[
\text{Differentiate term by term:}
\]
\[
\frac{d}{dx} (5x^3 + 4x^2 – 6x + 7) = 15x^2 + 8x – 6.
\]
\[
\text{Thus, the correct answer is } \boxed{15x^2 + 8x – 6}.
\]

More MCQs on Calculus

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