Question

${lim}_{x\to 0}\left(\frac{e^{x^2}-\cos x}{x^2}\right)$ is equal to

• A. $\frac{3}{2}$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. None of these

Answer 1

The correct option is A $\frac{3}{2}$

Given $\underset{x\to 0}{lim}\left(\frac{e^{x^2}-\cos x}{x^2}\right)$

$=\underset{x\to 0}{lim}\left(\frac{2e^{x^2}x+\sin x}{2x}\right)$

$=\underset{x\to 0}{lim}\left(\frac{2(2e^{x^2}{x}^2+e^{x^2})+\cos x}{2}\right)$

Putting the value of x=0

$=\frac{2(2e^{0^2}{.0}^2+e^{0^2})+\cos 0}{2}$

$=\frac{2(0+1)+1}{2}$

$=\frac{3}{2}$