Original JEE Main and CUET-style questions on limits, continuity and differentiability, with a worked solution and the usual trap for each. Tap an option, or type a numerical answer, to check it.
Learn the chapters first: Class 11, Limits and Derivatives · Class 12, Continuity and Differentiability.
\(\displaystyle\lim_{x \to 0} \frac{1 - \cos 6x}{x\sin x}\) equals
\(1 - \cos 6x = 2\sin^2 3x\), so the expression is \(2 \cdot 9 \cdot \left(\dfrac{\sin 3x}{3x}\right)^2 \cdot \dfrac{x}{\sin x} \to 18\).
Trap: Using \(1 - \cos 6x \approx \tfrac{6x^2}{2}\) instead of \(\tfrac{(6x)^2}{2}\) gives 3: square the whole angle.
\(f(x) = \begin{cases} kx^2 - 1, & x \le 3 \\ 2x + 5, & x > 3 \end{cases}\) is continuous at \(x = 3\). Then \(k\) equals
Left value \(9k - 1\), right-hand limit \(6 + 5 = 11\). Continuity: \(9k - 1 = 11\), so \(k = \dfrac{12}{9} = \dfrac43\).
Find \(\displaystyle\lim_{x \to 2} \frac{x^6 - 64}{x - 2}\).
\(\displaystyle\lim_{x \to a}\frac{x^n - a^n}{x - a} = na^{n-1}\). With \(n = 6\), \(a = 2\): \(6 \cdot 2^5 = 192\).
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