How to calculate $\lim_{x\to1}\left(\frac{1+\cos(\pi x)}{\tan^2(\pi x)}\right)^{\!x^2}$

How can I calculate this Limit without l'Hôpital's rule?

Calculate $$\lim_{x\to1}\left(\frac{1+\cos(\pi x)}{\tan^2(\pi x)}\right)^{\!x^2}$$

All I got is this:

$$\exp \left(\left(\lim \limits_{x\to1}x^2\right)\ln\left(\lim \limits_{x\to1}\frac{1+\cos(\pi x)}{\tan^2(\pi x)}\right)\right)=\exp\left(\ln \left(\lim \limits_{x\to1}\frac{1+\cos(\pi x)}{\tan^2(\pi x)}\right)\right)$$

then I split the $\tan$ into $\dfrac\sin\cos$ and I dont know how to continue.

• $\tan$?? What's a $\tan$, are you on a beach. :) – Shobhit Oct 13 '13 at 18:21
• $\dfrac{1+\cos (\theta)}{(\tan (\theta))^2}=\ldots = \dfrac{(\cos (\theta))^2}{1-\cos (\theta)}$ – Git Gud Oct 13 '13 at 18:23
• @Shobhit, it would be best to avoid multiple trivial edits to a post needing substantial editing, as you will interrupt someone attempting a full edit. – dfeuer Oct 13 '13 at 18:26
• @dfeuer I feel you. It's so annoying. – Git Gud Oct 13 '13 at 18:27
• @dfeuer an edit is an edit(trivial it may be), i am doing the best i can for i do not know how to fancy a question, like you. Do not take it the other way. – Shobhit Oct 13 '13 at 18:31

Setting $x=1-h$

$$\lim_{x\to1}\frac{1+\cos\pi x}{\tan^2\pi x}=\lim_{h\to0}\frac{1+\cos\{\pi(1-h)\}}{[\tan\{\pi(1-h)\}]^2}$$

$$=\lim_{h\to0}\frac{1-\cos\pi h}{\tan^2\pi h}\text{ as }\cos(\pi-y)=-\cos y,\tan(\pi-y)=-\tan y$$

$$=\lim_{h\to0}\frac{1-\cos\pi h}{\sin^2\pi h}\cdot \left(\lim_{h\to0}\ \cos\pi h\right)^2$$

$$=\lim_{h\to0}\frac{1-\cos\pi h}{(1-\cos\pi h)(1+\cos\pi h)}\cdot1$$

$$=\lim_{h\to0}\frac1{1+\cos\pi h}$$ as $1-\cos\pi h\ne0$ as $h\ne0$ as $h\to0$

$$=\cdots$$