Question Number 176388 by Matica last updated on 18/Sep/22
$$\:{find}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}+{sinx}}{{x}−{sinx}} \\ $$
Commented by Matica last updated on 18/Sep/22
$$\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{{x}+{sinx}}{{x}−{sinx}}\:\overset{?} {=}\:−\infty \\ $$
Answered by cortano1 last updated on 18/Sep/22
![lim_(x→0) ((1+cos x)/(1−cos x)) = lim_(x→0) (((1+cos x)^2 )/(sin^2 x)) = 4. [ lim_(x→0) (1/(sin x)) ]^2 = 4(∞)=∞](https://www.tinkutara.com/question/Q176390.png)
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} }{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$=\:\mathrm{4}.\:\left[\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{x}}\:\right]^{\mathrm{2}} =\:\mathrm{4}\left(\infty\right)=\infty\: \\ $$