Question

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

• A. $\frac{3}{2}$
• B. $-\frac{3}{2}$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

Answer: (A)

$\frac{3}{2}$

${lim}_{x\to 0}\frac{e^{x^2}-\cos x}{x^2}$
$=\underset{x\to 0}{lim}\frac{e^{x^2}-1+1-\cos x}{x^2}$
$=\underset{x\to 0}{lim}(\frac{e^{x^2}-1}{x^2}+\frac{1-\cos x}{x^2})$
$=1+\underset{x\to 0}{lim}\left(\frac{1-\cos x}{x^2}\right)$
$=1+\underset{x\to 0}{lim}\frac{2{\sin }^2\frac{x}{2}}{4\left(\frac{x^2}{4}\right)}$
$=\frac{3}{2}$