Question Number 116588 by bemath last updated on 05/Oct/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4x}}{\mathrm{2}\:\mathrm{cosec}\:\mathrm{x}\:\left(\mathrm{1}−\sqrt{\mathrm{cos}\:\mathrm{x}}\right)}\:=? \\ $$
Answered by bobhans last updated on 05/Oct/20
![lim_(x→0) ((4x.sin x)/(2(1−(√(1−2sin^2 ((x/2))))))) = lim_(x→0) ((4x.sin x)/(2(1−(1−((2((x^2 /4)))/2))))) = lim_(x→0) [((2x^2 )/(((x^2 /4))))] = 8](https://www.tinkutara.com/question/Q116589.png)
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4x}.\mathrm{sin}\:\mathrm{x}}{\mathrm{2}\left(\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\right)}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4x}.\mathrm{sin}\:\mathrm{x}}{\mathrm{2}\left(\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{2}\left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4}}\right)}{\mathrm{2}}\right)\right)}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{2x}^{\mathrm{2}} }{\left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4}}\right)}\right]\:=\:\mathrm{8} \\ $$
Answered by Dwaipayan Shikari last updated on 05/Oct/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4x}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{1}−\sqrt{\mathrm{cosx}}\right)}=\frac{\mathrm{2x}^{\mathrm{2}} }{\mathrm{1}−\mathrm{cosx}}\left(\mathrm{1}+\sqrt{\mathrm{cosx}}\right) \\ $$$$=\mathrm{2}.\frac{\mathrm{2x}^{\mathrm{2}} }{\mathrm{2sin}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}}=\mathrm{8} \\ $$