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Question Number 160609 by cortano last updated on 03/Dec/21

$$\underset{x\to \pi }{\lim }\left(\frac{\tan \mathrm{x}}{1+\cos \mathrm{x}}\right)=?$$

Answered by Ar Brandon last updated on 03/Dec/21

$$\mathcal{L}=\underset{x\to \pi }{\lim }\left(\frac{\tan x}{1+\cos x}\right),u=x-\pi$$$$=\underset{u\to 0}{\lim }\left(\frac{\tan (u+\pi )}{1+\cos (u+\pi )}\right)=\underset{u\to 0}{\lim }\left(\frac{\tan u}{1-\cos u}\right)$$$$=\underset{u\to 0}{\lim }\left(\frac{\frac{u}{1-\frac{u^2}{2}}}{1-(1-\frac{u^2}{2})}\right)=\underset{u\to 0}{\lim }(\frac{2u}{2-u^2}\cdot \frac{2}{u^2})$$$$=\underset{u\to 0}{\lim }\left(\frac{4}{2u-u^3}\right)\to \pm \mathrm{\infty }$$

Answered by alephzero last updated on 03/Dec/21

$$\underset{x\to \pi }{\lim }\left(\frac{\tan \mathrm{x}}{1+\cos \mathrm{x}}\right)=?$$$$Tobehonest,I^{\prime }min{6}^{th}grade.$$$$ButIcantry$$$$\underset{x\to \pi }{\lim }\frac{\tan x}{1+\cos x}=?$$$$\tan \pi =0$$$$\cos \pi =-1$$$$1+\cos \pi =0$$$$Thisequationmaylookweird,$$$$but{let}^{\prime }sremembertherulesof$$$$arithmetic.$$$$\mathrm{\forall }n\{n\in \mathbb{R}\mid \frac{n}{n}=1\}$$$$Then,$$$$\underset{x\to \pi }{\lim }\frac{\tan x}{1+\cos x}=\frac{\tan \pi }{1+\cos \pi }=\frac{0}{0}=1$$$$Or,if$$$$\underset{x\to 0}{\lim }\frac{n}{x}\to \mathrm{\infty },wheren\in \mathbb{R}$$$$then$$$$\underset{x\to \pi }{\lim }\frac{\tan x}{1+\cos x}=\underset{x\to 0}{\lim }\frac{0}{x}=\mathrm{\infty }$$$$Ifthisisfalse,{don}^{\prime }tmindme,$$$$I{don}^{\prime }tknow{L}^{\prime }H{\widehat{opital}}^{\prime }srule,$$$$and{I}^{\prime }monly12yearsold.$$$$$$

Answered by tounghoungko last updated on 03/Dec/21

$$\underset{x\to \pi }{\lim }\left(\frac{2\sin \frac{1}{2}\mathrm{x}\cos \frac{1}{2}\mathrm{x}}{2\cos {}^2\frac{1}{2}\mathrm{x}\cos \mathrm{x}}\right)=$$$$\underset{x\to \pi }{\lim }\left(\frac{\tan \frac{1}{2}\mathrm{x}}{\cos \mathrm{x}}\right)=-\mathrm{\infty }$$

Answered by Mathspace last updated on 03/Dec/21

$$f\left(x\right)=\frac{tanx}{1+cosx}$$$$changementx-\pi =tgive$$$$f\left(x\right)=f(t+\pi )=\frac{tan(t+\pi )}{1+cos(t+\pi )}(t\to 0)$$$$=\frac{tant}{1-cost}\sim \frac{t}{\frac{t^2}{2}}=\frac{2}{t}\to \mathrm{\infty }$$

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