Limits, Continuity & Differentiability
Differential Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
If $f(x) = \begin{cases} \frac{[(a-n)nx - \tan x]\sin nx}{x^2} & \text{at } x = 0 \\ 0 & \text{at } x = 0 \end{cases}$, where $n$ is a non-zero real number, and $f$ is continuous at $x = 0$, then $a$ is equal to:
0
$\frac{n}{n+1}$
$n$
$n + \frac{1}{n}$
Step-by-Step Solution
Key Concept: Separate the limit into manageable parts using Taylor expansions and properties of standard limits.
We evaluate $\lim_{x \to 0} \frac{(a-n)nx - \tan x) \sin nx}{x^2} = 0$ by expanding the numerator. Using Taylor series, we separate the limit as $\lim_{x \to 0} \left[(a-n)n - \frac{\tan x}{x}\right] \frac{\sin nx}{nx} = 0$. Since $\frac{\sin nx}{nx} \to 1$ as $x \to 0$, we need $(a-n)n - 1 = 0$, giving $a = \frac{1}{n} + n$.
Correct Answer: 4