Calculate the Given Limit without using L'Hôpital's and Expansion

Question

If $$\lim_{x \to 0} \frac{e^{\frac{-x^2}{2}}-\cos x}{x^3 \sin x}$$

Exsit then Calculate Above limit without using L'Hôpital's and Expansion.

My Approach:

$$l= \lim_{x \to 0} \frac{e^{\frac{-x^2}{2}}-\cos x}{x^3 \sin x}$$

$$l= \lim_{x \to 0} \frac{(e^{\frac{-x^2}{2}}-1)+(1-\cos x)}{x^3 \sin x}$$

$$l= \lim_{x \to 0} \frac{(e^\frac{-x^2}{2}-1)(\frac{-x^2}{2})}{\frac{-x^2}{2} \cdot (x^4) \cdot(\frac{\sin x}{x})}+\frac{(1-\cos x)(x^2)}{(x^2 ) \cdot \frac{(\sin x)}{x} \cdot x^4}$$

But I am Stuck Here. Any help will be Appreciated.

Answer 1

HINT

We have that

$$\frac{e^{\frac{-x^2}{2}}-\cos x}{x^3 \sin x} =\frac{e^{\frac{-x^2}{2}}-\cos x}{x^4} \frac x {\sin x}$$

and

$$\frac{e^{\frac{-x^2}{2}}-\cos x}{x^4}=\frac14 \frac{e^{\frac{-x^2}{2}}-1+\frac {x^2}2}{\frac{x^4}4}+ \frac{1-\frac {x^2}2-\cos x}{x^4}$$

then refer to

• Are all limits solvable without L'Hôpital Rule or Series Expansion