The value of $\underset{x\to 0}{lim} \frac{e^x-\cos 2x-x}{x^2},$ is 

Solution:

     $\underset{x\to 0}{lim} \frac{e^x-\cos 2x-x}{x^2}$

$\begin{array}{l}=\underset{x\to 0}{lim} \frac{\left(e^x-1-x\right)+(1-\cos 2x)}{x^2}\\ =\underset{x\to 0}{lim} \frac{\left(\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \right)+2{\sin }^{2}x}{x^2}\\ =\underset{x\to 0}{lim} \left\{\left(\frac{1}{2!}+\frac{x}{3!}+\dots \right)+2{\left(\frac{\sin x}{x}\right)}^2\right\}=\frac{1}{2!}+2\times 1^2=\frac{5}{2}\end{array}$