Question

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

• A. $0$
• B. $\frac{3}{2}$
• C. $-1$
• D. $doesnotexist$

Answer 1

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

Now,

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

Now applying L'Hospital's rule we get,

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

$=\underset{x\to 0}{lim}\frac{2e^{x^2}+4x^2{e}^{x^2}+\cos x}{2}$

$=\frac{3}{2}$.