Find limit of $\lim_{x \to 0^+}\frac{\sin^2(x)}{e^{-1/x}}$

Question

I’m asked to find the following limit: $$\lim_{x \to 0^+}\frac{\sin^2(x)}{e^{-1/x}}$$

First I thought about changing $e$ in its Taylor series, but I don’t see what I could do next. I also rewrote sine squared as $1 - \cos^2$, but that doesn’t help too. And I don’t see why I should take the Taylor series of $\sin$ given that it is squared, that would be a mess. L’Hôpital’s rule doesn’t help either because we get (..)/$x^2$ then (...)/$x^4$ etc. So I don’t see what I could do next.

Thanks for your help !

Answer 1

$$\lim_{x\to 0^+}\frac{\sin^2x}{e^{-1/x}}=\lim_{x\to 0^+}\left(\left(\frac{\sin x}{x}\right)^2\cdot \frac{x^2}{e^{-1/x}}\right) = \lim_{x\to 0^+}\frac{x^2}{e^{-1/x}} = \lim_{x\to + \infty} \frac{e^x}{x^2} = +\infty$$

Answer 2

Hints:

1. Get rid of $\sin^2 x$ by using $\frac{\sin x}{x}\to 1$ as $x\to 0$.
2. After that change the variable $t=1/x$.

Answer 3

Let $t=1/x$, then the limit you want to compute changes to $$\lim_{t\to+\infty}\frac{e^t}{t^2}=+\infty.$$