I need help calculating this limit: \(\displaystyle\lim_{{{x}\to\pi}}{\frac{{{\cos{{2}}}{x}+{2}{\left(\pi-{x}\right)}{\sin{{x}}}-{1}}}{{{\left({\cos{{2}}}{x}+{\cos{{x}}}\right)}^{{2}}}}}\) I tried

Question

I need help calculating this limit:

lim_{x \to π} \frac{\cos 2 x + 2 (π - x) \sin x - 1}{{(\cos 2 x + \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 + π

Thus,

lim_{x \to π} \frac{\cos 2 x + 2 (π - x) \sin x - 1}{{(\cos 2 x + \cos x)}^{2}} = lim_{t \to 0} \frac{\cos 2 t + 2 t \sin t - 1}{{(\cos 2 t - \cos t)}^{2}} =

= lim_{t \to 0} \frac{2 t \sin t - 2 \sin^{2} t}{{(1 - \cos t)}^{2} {(2 \cos t + 1)}^{2}} = \frac{2}{9} lim_{t \to 0} \frac{t \sin t - \sin^{2} t}{4 \sin^{4} \frac{t}{2}} =

= \frac{8}{9} lim_{t \to 0} \frac{t \sin t - \sin^{2} t}{t^{4}} = \frac{8}{9} lim_{t \to 0} \frac{t - \sin t}{t^{3}} =

= \frac{8}{9} lim_{t \to 0} \frac{1 - \cos {t}}{3 t^{2}} = \frac{8}{9} lim_{t \to 0} \frac{2 \sin^{2} \frac{t}{2}}{3 t^{2}} = \frac{8}{9} lim_{t \to 0} \frac{2 \cdot \frac{t^{2}}{4}}{3 t^{2}} = \frac{4}{27}

Answer 2

Hint: breaks it like this

lim_{x \to π} \frac{\cos 2 x + 2 (π - x) \sin x - 1}{{(\cos 2 x + \cos x)}^{2}}

= lim_{x \to π} (\frac{\cos 2 x - 1}{{(π - x)}^{2}} + 2 \frac{\sin x}{(π - x)}) \cdot lim_{x \to π} {(\frac{π - x}{\cos 2 x + \cos x})}^{2}

Then use the definition of derivative of a function at

x = π

Answers (2)