Mathematics - Limits Question with Solution | TestHub

Given, $\underset{x\to 0}{\lim }\frac{e^{2\left|\sin x\right|}-2\left|\sin x\right|-1}{x^2}$

Now, using the expansion of $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+........\infty$ we get,

$=\underset{x\to 0}{\lim }\frac{\left(1+\frac{\left|2\sin x\right|}{1!}+\frac{{\left|2\sin x\right|}^2}{2!}+.....\infty \right)-\left|2\sin x\right|-1}{x^2}$

$=\underset{x\to 0}{\lim }\frac{\frac{{\left|2\sin x\right|}^2}{2!}+\frac{{\left|2\sin x\right|}^3}{3!}.....\infty }{x^2}$

$=\underset{x\to 0}{\lim }\frac{2{\sin }^{2}x+\frac{{\left|2\sin x\right|}^3}{3!}.....\infty }{x^2}$

$=2 \left\{\text{as} \underset{x\to 0}{\lim }\frac{{\sin }^{2}x}{x^2}=1\text{ and other terms will become zero}\right\}$