Question

If $L=\underset{x\to 0}{lim}\frac{e^{-x^2/2}-cosx}{x^{3}sinx}$, then the value of $\frac{1}{3L}$ is

Answer 1

$L={lim}_{x\to 0}\frac{e^{-x^2/2}-\cos x}{x^3}\times \frac{x}{\sin x}\times \frac{1}{x}$

$={lim}_{x\to 0}\frac{e^{-x^2/2}-\cos x}{x^4}\frac{0}{0}$

$={lim}_{x\to 0}\frac{-x{e}^{-x^2/2}+\sin x}{4x^3}$

$={lim}_{x\to 0}\frac{(-x)(1+\frac{(-x^2/2)}{1!})+(x-\frac{x-x^3}{3!})}{4x^3}$

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+....,e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}.....$$={lim}_{x\to 0}\frac{-x+\frac{x^3}{2}+x-\frac{x^3}{6}}{4x^3}$

$L={lim}_{x\to 0}\frac{2x^3}{6\times 4\times x^3}=\frac{1}{12}$

$\frac{1}{3L}=\frac{1}{3}\times \frac{12}{1}=4$