How to calculate this limit to $\pi$?

Question

I need help calculating this limit:

$$ \lim_{x\to \pi}\frac{\cos2x+2(\pi-x)\sin x-1}{(\cos2x+\cos x)^2} $$ I tried to use L'Hospital's rule, but didn't get any result. Can you help me please?

Answer 1

Let $x=t+\pi$.

Thus, $$ \lim_{x\to \pi}\frac{\cos2x+2(\pi-x)\sin x-1}{(\cos2x+\cos x)^2}=\lim_{t\rightarrow0}\frac{\cos2t+2t\sin{t}-1}{(\cos2t-\cos{t})^2}=$$ $$=\lim_{t\rightarrow0}\frac{2t\sin{t}-2\sin^2t}{(1-\cos{t})^2(2\cos{t}+1)^2}=\frac{2}{9}\lim_{t\rightarrow0}\frac{t\sin{t}-\sin^2t}{4\sin^4\frac{t}{2}}=$$ $$=\frac{8}{9}\lim_{t\rightarrow0}\frac{t\sin{t}-\sin^2{t}}{t^4}=\frac{8}{9}\lim_{t\rightarrow0}\frac{t-\sin{t}}{t^3}=$$ $$=\frac{8}{9}\lim_{t\rightarrow0}\frac{1-\cos{t}}{3t^2}=\frac{8}{9}\lim_{t\rightarrow0}\frac{2\sin^2\frac{t}{2}}{3t^2}=\frac{8}{9}\lim_{t\rightarrow0}\frac{2\cdot\frac{t^2}{4}}{3t^2}=\frac{4}{27}$$

Answer 2

Hint: breaks it like this $$\lim_{x\to \pi}\frac{\cos2x+2(\pi-x)\sin x-1}{(\cos2x+\cos x)^2} \\=\lim_{x\to \pi}\left(\frac{\cos2x-1}{(\pi-x)^2 }+2\frac{\sin x}{(\pi-x)}\right)\cdot\lim_{x\to \pi}\left(\frac{\pi-x}{\cos2x+\cos x}\right)^2$$

Then use the definition of derivative of a function at $x=\pi.$