Limit for $x\to\pi$ for the trigonometric function $(1+\cos x)/\tan^2x$

Question

$\displaystyle \lim_{x\to\pi}\frac{1+\cos x}{\tan^2x}$

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.

Answer 1

HINT

We have that

$$\frac{1+\cos x}{\tan^2x}=\frac{\cos^2x(1+\cos x)}{\sin^2x}=\frac{\cos^2x(1+\cos x)}{1-\cos^2x}=\frac{\cos^2x}{1-\cos x}$$

which is not an indeterminate form.

Answer 2

Hint: $1 + \tan^2(x) = 1/\cos^2(x)$ (divide trig identity by $cos^2(x)$)

Your expression turns into

$\lim_{x \to\pi}\frac{(cos(x)+1)*cos^2(x)}{1 - cos^2(x)}$

Can you progress from there?

Answer 3

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: $$ \frac{1+\cos x}{\tan^2x}= \frac{(1+\cos x)(1-\cos x)}{\sin^2x}\frac{\cos^2x}{1-\cos x} $$

Answer 4

Hint 1

Note, that

• $\lim\limits_{x\to \pi}\tan x= \lim\limits_{x\to 0}\tan x$
• $\lim\limits_{x\to \pi}\cos x= \lim\limits_{x\to 0}-\cos x$

Hint 2

$1-\cos x = 2 \sin^2 \frac{x}{2}$

Hint 3

$\lim\limits_{\alpha \to 0} \frac{\sin \alpha}{\alpha}=1$