find $\lim_{x \to {0^-}}(e^{1/x})\cos x$ without using l'Hopital's rule

Question

I would separate this limit but $1/x$ is an indeterminate form, which is where I'm stuck.

Answer 1

$\cos(x) \to 1$ and $\lim_{x \to 0^{-}} e^{1/x} = 0$ so the limit is their product or $0$.

Answer 2

Try subbing in $u = 1/x$. After doing this, the limit now goes to $-\infty$.

$\lim e^u\cos(1/u)$

Then, we can use squeeze theorem:

$-1 < \cos(1/u) < 1$

$-e^u < e^u\cos(1/u) < e^u$

Remember that $e^u$ approaches $0$ as u goes to $-\infty$. Subbing in $-\infty$ squeezes $e^u\cos(1/u)$ into $0$. The limit $ = 0$