I started with the Taylor expansions of $\cos(1/x)$ and $e^x$
i.e
$\cos(1/x) : [ 1 - \frac{1}{2x^2} + o(\frac{1}{x^4}) ]$
and
$e^x : ( 1 + x + \frac{x^2}{2} + o(x^3) )$
I realise $e^x$ goes to $1$ as $x \to 0$ so can we ignore the $e^x$ term and focus only on the $\cos(1/x)$ term? How would you proceed from there on?
On
In the Limit Toolbox we have: If $\lim_{x\to a}U(x)=P\in \Bbb R$ and $\lim_{x\to a}V(x)=Q\in \Bbb R$ then $\lim_{x\to a}U(x)V(x)=PQ.$
With $a=0,$ let $U(x)=e^x\cos (1/x)$ for $x\ne 0$ and let $V(x)=e^{-x}.$ We have $\lim_{x\to a}V(x)=1=Q\in \Bbb R.$
IF $\lim_{x\to a}U(x)=P\in \Bbb R$ THEN $\lim_{x\to 0}\cos (1/x)=\lim_{x\to a}U(x)V(x)=PQ\in \Bbb R.$
Does $\cos (1/x)$ have a limit as $x\to 0?$
The limits along the sequences $(\frac 1 {\frac {\pi } 2+2n \pi})$ and $(\frac 1 {2n \pi})$ are $0$ and $1$ respectively. Hence the limit does not exist.
[$\cos (\frac {\pi } 2+2n \pi)=0$ for all $n$, $\cos ( 2n \pi)=1$ for all $n$ and $e^{x}$ part can be ignored as you have already observed. For existence of the limit it is necessary that limit along all sequences converging to $)$ are the same].