Home › Forums › Math questions › Calculus questions

Trigonometric Limits

Posted December 5th, 2009 by alpha
in

Find:
$ \mathop{\lim}\limits_{x\to \pi}\frac{\sqrt{5+cosx}-2}{(\pi-x)^2} $

‹ Trigonometric Limits Limits ›

Comments

answer

On December 5th, 2009 Structure says:
$$\lim_{x\to \pi}\frac{\sqrt{5+cosx}-2}{(\pi-x)^2} = $$
$$\lim_{x \to \pi}\frac{5+\cos x-4}<br /> {(\sqrt{5+\cos x}+2)(\pi-x)^2} = \lim_{x\to \pi}\frac{1-cos(\pi-x)}{(\pi-x)^2}\frac{1}{\sqrt{5+cosx}+2}=\frac{1}{4}\lim_{y\to 0}\frac{1-\cos y}{y^2}=\frac{1}{8}$$

Back to top