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BITSAT Limits and Continuity — practice questions

Explore 110 free Limits and Continuity questions in BITSAT Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Limits and Continuity questions and exercises with answers

Showing questions 1–30 of 110.

Evaluate $\displaystyle\lim_{x \to 0} \dfrac{\sin 3x}{x}$.Read the answer and worked solution →Evaluate $\displaystyle\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2}$.Read the answer and worked solution →Evaluate $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos 2x}{x^2}$.Read the answer and worked solution →Evaluate $\displaystyle\lim_{x \to \infty} \left(1 + \dfrac{2}{x}\right)^x$.Read the answer and worked solution →Consider $f(x) = |x|$ at $x = 0$. Which statement is correct?Read the answer and worked solution →$\sin(180° - \theta)$ equals:Read the answer and worked solution →$\sin^2\theta + \cos^2\theta$ equals:Read the answer and worked solution →$\sin(A + B) = $Read the answer and worked solution →sin(0°) equals:Read the answer and worked solution →cos(90°) equals:Read the answer and worked solution →sin²θ + cos²θ equals:Read the answer and worked solution →tan(45°) equals:Read the answer and worked solution →In which quadrant is sin θ positive and cos θ negative?Read the answer and worked solution →lim_(x→2) (x² - 4)/(x - 2) =Read the answer and worked solution →lim_(x→0) sin(x)/x =Read the answer and worked solution →For limit to EXIST at x = a, left-hand and right-hand limits must:Read the answer and worked solution →A function f is continuous at x = a if:Read the answer and worked solution →d/dx (xⁿ) = ?Read the answer and worked solution →lim_(x→0) (1 - cos x)/x² =Read the answer and worked solution →lim_(x→∞) (1 + 1/x)^x =Read the answer and worked solution →lim_(x→0) (e^x - 1)/x =Read the answer and worked solution →lim_(x→0) (sin 3x)/(sin 5x) =Read the answer and worked solution →d/dx (sin x) =Read the answer and worked solution →d/dx (eˣ) =Read the answer and worked solution →d/dx (ln x) =Read the answer and worked solution →Find d/dx (x² sin x) by product rule:Read the answer and worked solution →By chain rule, d/dx [sin(x²)] =Read the answer and worked solution →lim_(x→∞) (3x² - 5)/(7x² + 2) =Read the answer and worked solution →l'Hôpital's rule applies to limits of form:Read the answer and worked solution →Find dy/dx by implicit differentiation for x² + y² = 25:Read the answer and worked solution →

Worked practice questions

Evaluate $\displaystyle\lim_{x \to 0} \dfrac{\sin 3x}{x}$.

  1. $1$
  2. $3$
  3. $\infty$
  4. $0$
Show answer and solution

Correct answer: B. $3$

Using the standard limit $\lim_{u\to 0} \dfrac{\sin u}{u} = 1$, write $\dfrac{\sin 3x}{x} = 3 \cdot \dfrac{\sin 3x}{3x}$. As $x \to 0$, the second factor $\to 1$, giving $3$.

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Evaluate $\displaystyle\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2}$.

  1. The limit does not exist
  2. $4$
  3. $2$
  4. $0$
Show answer and solution

Correct answer: B. $4$

Factor the numerator: $x^2 - 4 = (x-2)(x+2)$. Cancelling, $\dfrac{(x-2)(x+2)}{x-2} = x + 2$ for $x \neq 2$. Limit = $4$. Option D is the trap if you stop at the $0/0$ form without factoring.

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Evaluate $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos 2x}{x^2}$.

  1. $2$
  2. $4$
  3. $0$
  4. $1$
Show answer and solution

Correct answer: A. $2$

Use $1 - \cos 2x = 2\sin^2 x$. Then $\dfrac{2\sin^2 x}{x^2} = 2\left(\dfrac{\sin x}{x}\right)^2 \to 2 \cdot 1 = 2$ as $x \to 0$. Alternatively, expand $\cos 2x \approx 1 - 2x^2$ to leading order, giving $\dfrac{2x^2}{x^2} = 2$.

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