Question

Evaluate $\underset{x\to 0}{lim}\frac{x\tan x}{(1-\cos x)}$

Answer 1

${lim}_{x\to 0}\frac{x\tan x}{1-\cos x}$

$={lim}_{x\to 0}\frac{x.{\sec }^{2}x+\tan x}{\sin x}$

$={lim}_{x\to 0}\frac{x.2{\sec }^{2}x\tan x+{\sec }^{2}x\tan x}{\cos x}$

$=1.$

Hence, the answer is $1.$