Evaluate $\lim\limits_{x \to 0} \left(\frac{x}{\sin x }\right)^{1/x^2}$.

I think I am supposed to use L'Hopital's rule but I cannot seem to get it in the form of $\frac 0 0$ or $\pm\frac\infty\infty$. Here is what I have done so far.

Take the left and right hand limits of the numerator:

$\lim_{x\to 0-} x^{1/x^2} =\lim_{x\to 0-} e^{1/x^2 \ln x} = e^{\infty \cdot (-\infty)}$ but this is not of either form mentioned. Similarly we get for the right hand limit. Any help is appreciated thanks!

Solution 1:

You have $1^{\infty}$ form. so use this :-

$$\lim_{x\to a}f(x)^{g(x)} = \exp(\lim_{x\to a} (f(x)-1)g(x))$$ when $\lim_{x\to a}f(x)^{g(x)}$ is $1^{\infty}$ form.

Here is a proof for the above

So you have :-

$$\exp(\lim_{x\to 0}\frac{x-\sin(x)}{x^{2}\sin(x)})=\\\exp(\lim_{x\to 0}\frac{x-\sin(x)}{x^{3}}\frac{x}{\sin(x)})=\\\exp(\lim_{x\to 0}\frac{x-\sin(x)}{x^{3}})$$.

Now use L'Hospital:-

$$\exp(\lim_{x\to 0}\frac{x-\sin(x)}{x^{3}})=\exp(\lim_{x\to 0}\frac{1-\cos(x)}{3x^{2}})=\\\exp(\lim_{x\to 0}\frac{\sin(x)}{6x})=\exp(\frac{1}{6})$$.

Giving you the final answer as $$e^{\frac{1}{6}}$$