Question

If $\underset{x\to 0}{lim}\frac{\{(a-n)nx-\tan x\}\sin nx}{x^2}=0,n\ne 0$ then $a$ is equal to

• A. $0$
• B. $1+\frac{1}{n}$
• C. $n$
• D. $n+\frac{1}{n}$

Answer 1

The correct option is D $n+\frac{1}{n}$

$\underset{x\to 0}{lim}\frac{((a-n)nx-\tan x)\sin nx}{x^2}=0$

Applying L-Hospital's rule

$\Rightarrow \underset{x\to 0}{lim}\frac{((a-n)n-{\sec }^{2}2x)\sin nx-n\cos nx(n(n-a)x+\tan x)}{2x}=0$

Again applying L-Hospital's rule

$\Rightarrow \underset{x\to 0}{lim}\frac{\sin nx(n^{3}x(n-a)+\tan x(n^2-2{\sec }^{2}x))-2n\cos nx(n(n-a)+{\sec }^{2}x)}{2}=0$

$\Rightarrow \frac{-n(1-an+n^2)}{2}=0\Rightarrow a=n+\frac{1}{n}$