UPSC NDA Math Model Paper 9

$\lim\limits_{x\to 0}\frac{\tan x - x}{x^{2} \tan x}=\frac{0}{0}$ is an indeterminate form. Let us simplify and use the L'Hospital's Rule. $\lim\limits_{x\to 0}\frac{\tan x - x}{x^{2} \tan x}$ $=\lim\limits_{x\to 0}\left[\frac{\tan x - x}{x^{3}} \times \frac{x}{\tan x}\right]$ We know that $\lim\limits_{x\to 0}\frac{x}{\tan x}=1$, but $\lim\limits_{x\to 0}\frac{\tan x - x}{x^{3}}$ is still an indeterminate form, so we use L'Hospital's Rule: $\lim\limits_{x\to 0}\frac{\tan x - x}{x^{3}} = \lim\limits_{x\to 0}\frac{\sec^{2} x - 1}{3 x^{2}}$ , which is still an indeterminate form, so we use L'Hospital's Rule again: $\lim\limits_{x\to 0}\frac{\sec^{2} x - 1}{3 x^{2}} = \lim\limits_{x\to 0}\frac{2\sec x (\sec x \tan x)}{6 x}$ $=\lim\limits_{x\to 0}\frac{\sec^{2} x \tan x}{3 x}$ , which is still an indeterminate form, so we use L'Hospital's Rule again: $\lim\limits_{x\to 0}\frac{\sec^{2} x \tan x}{3 x}$ $=\lim\limits_{x\to 0}\frac{\sec^{2} x \sec^{2} x + \tan x [2 \sec x (\sec x \tan x)]}{3}$ $=\frac{1}{3}$ $\therefore \lim\limits_{x\to 0}\frac{\tan x - x}{x^{2} \tan x} = 1 \times \frac{1}{3} = \frac{1}{3}.$