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Question Number 86031 by arcana last updated on 26/Mar/20

$$\underset{x\to 0}{\lim }\frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin {}^{2}x}=$$

Commented by arcana last updated on 26/Mar/20

$$theansweris\frac{1}{4\sqrt{2}}$$

Commented by mathmax by abdo last updated on 26/Mar/20

$$idontthinkchowyourwork...$$

Commented by abdomathmax last updated on 27/Mar/20

$$forgive{u}^{\left(2\right)}\left(0\right)=\frac{2\sqrt{2}}{8}=\frac{\sqrt{2}}{4}\Rightarrow lim(...)=\frac{\sqrt{2}}{8}$$

Commented by john santu last updated on 26/Mar/20

$$notcorrect$$

Commented by abdomathmax last updated on 27/Mar/20

$$letverifybyhospitaltheorem$$$$u\left(x\right)=\sqrt{2}-\sqrt{1+cosx}\Rightarrow u^{{}^{\prime }}\left(x\right)=\frac{sinx}{2\sqrt{1+cosx}}$$$$u^{\left(2\right)}\left(x\right)=\frac{2cosx\sqrt{1+cosx}-sinx\times \frac{-sinx}{\sqrt{1+cosx}}}{4(1+cosx)}\Rightarrow$$$$u^{\left(2\right)}\left(0\right)=\frac{2\sqrt{2}}{4}$$$$v\left(x\right)={sin}^{2}x\Rightarrow v^{{}^{\prime }}\left(x\right)=2sinxcosxand$$$$v^{\left(2\right)}\left(x\right)=2{cos}^{2}x-2{sin}^{2}x\Rightarrow v^{\left(2\right)}\left(0\right)=2\Rightarrow$$$${lim}_{x\to 0}\frac{\sqrt{2}-\sqrt{1+cosx}}{{sin}^{2}x}=\frac{\sqrt{2}}{4}$$

Commented by arcana last updated on 27/Mar/20

$$Ithink{it}^{\prime }senough$$$$\underset{x\to 0}{\lim }\frac{\sqrt{2}-\sqrt{1+\cos (x})}{\sin {}^2\left(x\right)}=\underset{x\to 0}{\lim }\frac{\frac{\sin \left(x\right)}{2\sqrt{1+\cos \left(x\right)}}}{2\sin \left(x\right)\cos \left(x\right)}$$$$=\underset{x\to 0}{\lim }\frac{1}{4\cos \left(x\right)\sqrt{1+\cos \left(x\right)}}=\frac{1}{4\sqrt{2}}$$

Answered by john santu last updated on 26/Mar/20

$$\underset{x\to 0}{\lim }\frac{2-(1+\cos x)}{\sin {}^{2}x}\times \underset{x\to 0}{\lim }\frac{1}{\sqrt{2}+\sqrt{1+\cos x}}$$$$=\underset{x\to 0}{\lim }\frac{1-\cos x}{\sin {}^{2}x}\times \frac{1}{2\sqrt{2}}$$$$=\frac{\sqrt{2}}{4}\times \underset{x\to 0}{\lim }\frac{2\sin {}^2\left(\frac{1}{2}x\right)}{\sin {}^{2}x}$$$$=\frac{\sqrt{2}}{4}\times \frac{\frac{1}{2}}{1}=\frac{\sqrt{2}}{8}$$$$=\frac{1}{4\sqrt{2}}$$

Commented by arcana last updated on 27/Mar/20

$$perfect!buthowyoucalculatedthispart$$$$\underset{x\to 0}{\lim }\frac{2\sin {}^2\left(\frac{x}{2}\right)}{\sin {}^2\left(x\right)}=\frac{\frac{1}{2}}{1}$$$$i{don}^{\prime }tunderstndyourmethod.$$$$Didyouuse\underset{x\to 0}{\lim sin}\left(x\right)/x=1?$$

Commented by john santu last updated on 27/Mar/20

$$yes.$$$$$$

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