$\displaystyle \lim_{x\to\pi}\frac{1+\cos x}{\tan^2x}$
How would I be able to solve this without using L'Hôpital's rule?
I have tried everything with normal identity manipulation etc and I have finally come to the decision it must be done somehow with the sandwich theorem but can't exactly see how.
I struggle to apply The squeeze/sandwich/ pinch theorem on limits with a trig in the numerator.
Best Answer
Note that $\cos(\pi-t)=-\cos t$ and $\tan(\pi-t)=-\tan t$, so the limit becomes, with $x=\pi-t$, $$ \lim_{t\to0}\frac{1-\cos t}{\tan^2t}= \lim_{t\to0}\frac{(1-\cos t)(1+\cos t)}{\sin^2t}\frac{\cos^2t}{1+\cos t} $$
Now, having seen this, we could do the trick directly: $$ \frac{1+\cos x}{\tan^2x}= \frac{(1+\cos x)(1-\cos x)}{\sin^2x}\frac{\cos^2x}{1-\cos x} $$