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 117687 by bemath last updated on 13/Oct/20

$$\underset{x\to 0}{\lim }\frac{{(\ln \left(\cosh \mathrm{x}\right)-\ln \left(\cos \mathrm{x}\right))}^2}{\sqrt{\cosh \mathrm{x}}+\sqrt{\cos \mathrm{x}}-2}=?$$

Answered by 1549442205PVT last updated on 13/Oct/20

$$\underset{\mathrm{x}\to 0}{\lim }\frac{{(\ln \left(\cosh \mathrm{x}\right)-\ln \left(\cos \mathrm{x}\right))}^2}{\sqrt{\cosh \mathrm{x}}+\sqrt{\cos \mathrm{x}}-2}{=}^{\frac{0}{0}}$$$$\underset{{\mathrm{L}}^{\prime }\mathrm{Hopital}}{=}\underset{\mathrm{x}\to 0}{\lim }\frac{{(\ln \frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}-\ln \left(\mathrm{cosx}\right))}^2}{\sqrt{\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}}+\sqrt{\mathrm{cosx}}-2}{=}^{\frac{0}{0}}$$$$\underset{\mathrm{x}\to 0}{=\lim }\frac{2(\ln \frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}-\ln \left(\mathrm{cosx}\right))\times (\frac{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}+\mathrm{tanx})}{\frac{\frac{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}}{2}}{\sqrt{2({\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}})}}+\frac{-\mathrm{sinx}}{2\sqrt{\mathrm{cosx}}}}$$$$\underset{\mathrm{x}\to 0}{{=}^{\frac{0}{0}}\lim }\frac{2(\ln \frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}-\ln \left(\mathrm{cosx}\right))(\frac{4}{{({\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}})}^2}+(1+{\tan }^2\mathrm{x}))+2{(\frac{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}+\mathrm{tanx})}^2}{\frac{({\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}})\sqrt{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}-({\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}})\frac{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}}{2\sqrt{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}}}{2\sqrt{2}({\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}})}}$$$$=\frac{0}{\frac{2\sqrt{2}}{4\sqrt{2}}}=0$$$$\mathrm{Thus},\mathrm{final}\mathrm{result}\mathrm{I}=0$$

Commented by bemath last updated on 13/Oct/20

$$\mathrm{gave}\mathrm{kudos}$$

Answered by Dwaipayan Shikari last updated on 13/Oct/20

$$\underset{x\to 0}{\lim }\frac{{(log(e^x+e^{-x})-log(e^{ix}+e^{-ix}))}^2}{\frac{1}{\sqrt{2}}(\sqrt{e^x+e^{-x}}+\sqrt{e^{ix}+e^{-ix}})-2}$$$$\underset{x\to 0}{\lim }\sqrt{2}\frac{{(-x+log(e^{2x}+1)+ix-log(e^{2ix}+1))}^2}{e^{-\frac{x}{2}}\sqrt{e^{2x}+1}+e^{-i\frac{x}{2}}\sqrt{e^{2ix}+1}-2\sqrt{2}}$$$$\underset{x\to 0}{\lim }\sqrt{2}\frac{{(-x+ix)}^2}{(-\frac{x}{2}-i\frac{x}{2})-2\sqrt{2}}log(e^{2x}+1)=log(e^{2ix}+1)Asx\to 0$$$$And{e}^{-\frac{x}{2}}=-\frac{x}{2}and{e}^{-i\frac{x}{2}}=-i\frac{x}{2}Asx\to 0$$$$And\sqrt{e^{2ix}+1}=\sqrt{e^{2x}+1}Asx\to 0$$$$\underset{x\to 0}{\lim 2}\sqrt{2}\frac{x^2{(1-i)}^2}{-x(1+i)-4\sqrt{2}}=0$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com