Evaluate $\lim_{x \to \ \pi} \left( \frac{\cos(x)}{\sin^2(x)}+\cot^2(x)\right)$ without L'hopital's rule or series

Question

I'm trying to evaluate this limit:

$$\displaystyle{\lim_{x \to \ \pi}} \bigg( \frac{\cos(x)}{\sin^2(x)}+\cot^2(x)\bigg)$$

However I simply can't figure it out. Initially I tried to write the limit as:

$\displaystyle{\lim_{x \to \ \pi}} \bigg( \frac{\cos(x)}{\sin^2(x)}+\frac{\cos^2(x)}{\sin^2(x)}\bigg)$

and then tried combining it somehow. I then tried to rewrite the denominator as $\sin^2(x)=1-\cos^2(x)$ and then tried to multiply by the conjugate but to no avail.

Does anyone have any idea as to how I should approach this? I really am clueless at this point. Thanks!

Answer 1

You were on the right track! $$\begin{align} \lim_{x\to \pi }\frac{\cos x(1+\cos x)}{1-\cos^2 x} &= \lim_{x \to \pi}\frac{\cos x}{1-\cos(x)}\\ &= -\frac{1}{2} \end{align}$$

Answer 2

Let \begin{align} \mathrm L&={\lim_{x \to \ \pi}} \left( \frac{\cos(x)}{\sin^2(x)}+\cot^2(x)\right)\\ &=\lim_{x \to \pi} \frac{\cos x + \cos^2 x}{\sin^2 x}\\&=\lim_{x \to \pi} \frac{\cos x(1+ \cos x)}{1-\cos^2 x}\\&=\lim_{x \to \pi} \frac{\cos x(1+ \cos x)}{(1-\cos x )(1+\cos x)}\\&=\lim_{x \to \pi} \frac{\cos x}{(1-\cos x )}\\&=\frac{-1}{1-(-1)}\\&=-\frac 12 \end{align}

Answer 3

$$\displaystyle{\lim_{x \to \ \pi}} \bigg( \frac{\cos(x)}{\sin^2(x)}+\cot^2(x)\bigg) =\displaystyle{\lim_{x \to \ 0}} \bigg( \frac{-\cos(x)}{\sin^2(x)}+\cot^2(x)\bigg) =\lim_{x\to 0 }\frac{\cos x(-1+\cos x)}{1-\cos^2 x} = \lim_{x \to 0}-\frac{\cos x}{1+\cos(x)}\\ = -\frac{1}{2}$$ Given that $\cos( x-\pi)=-\cos x$ and that $\sin (x-\pi)=\sin x$