Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 141627 by cherokeesay last updated on 21/May/21

Answered by MJS_new last updated on 21/May/21

$$\frac{\sin 2x+\sin 4x}{1+\cos 2x+\cos 4x}=\frac{-2\sin x\cos x(1-{4\cos }^{2}x)}{1+{2\cos }^{2}x(1-{4\sin }^{2}x)}=$$$$[\sin x=\frac{\tan x}{\sqrt{1+{\tan }^{2}x}}\wedge \cos x=\frac{1}{\sqrt{1+{\tan }^{2}x}}]$$$$=\frac{2\tan x}{1-{\tan }^{2}x}$$$$\Rightarrow$$$$\underset{x\to \frac{\pi }{3}}{\lim }\frac{\sin 2x+\sin 4x}{1+\cos 2x+\cos 4x}=\underset{x\to \frac{\pi }{3}}{\lim }\frac{2\tan x}{1-{\tan }^{2}x}=-\sqrt{3}$$

Commented by cherokeesay last updated on 21/May/21

$$thankyousir.$$

Commented by MJS_new last updated on 21/May/21

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcome}$$

Answered by bramlexs22 last updated on 22/May/21

$$\underset{x\to \frac{\pi }{3}}{\lim }\frac{\sin 2x\cancel{(1+2\cos 2x)}}{\cos 2x\cancel{(1+2\cos 2x)}}$$$$=\underset{x\to \frac{\pi }{3}}{\lim }\tan 2x=\tan \left(\frac{2\pi }{3}\right)$$$$=-\tan 60°=-\sqrt{3}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com