Trigonometric Limits

Posted December 9th, 2009 by alpha

Find: $\mathop{\lim}\limits_{n \to \frac{\pi}{2}}\frac{cotx-cosx}{(\pi-2x)^3}$
You have to take

$y=\frac{\pi}{2}-x$

$\cot x=\cot (\frac{\pi}{2}-y)=\tan y$

$\cos x=\cos (\frac{\pi}{2}-y)=\sin y$

$\lim_{y \to 0}\frac{\tan y-\sin y}{8y^3}=\frac{1}{8}\lim_{y \to 0}\frac{\tan y}{y}\lim_{y \to 0}\frac{1-\cos y}{y^2}=\frac{1}{16}$

Comments

On December 11th, 2009 Structure says:

Find: $\mathop{\lim}\limits_{n \to \frac{\pi}{2}}\frac{cotx-cosx}{(\pi-2x)^3}$
You have to take

$y=\frac{\pi}{2}-x$

$\cot x=\cot (\frac{\pi}{2}-y)=\tan y$

$\cos x=\cos (\frac{\pi}{2}-y)=\sin y$

$\lim_{y \to 0}\frac{\tan y-\sin y}{8y^3}=\frac{1}{8}\lim_{y \to 0}\frac{\tan y}{y}\lim_{y \to 0}\frac{1-\cos y}{y^2}=\frac{1}{16}$