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 112866 by bemath last updated on 10/Sep/20

Answered by bobhans last updated on 10/Sep/20

$$\left(7\right)\mathrm{L}=\underset{x\to 0}{\lim }{\left(\frac{1}{\mathrm{x}}\right)}^{\tan \mathrm{x}}$$$$\mathrm{L}={\mathrm{e}}^{\underset{x\to 0}{\lim tan}\mathrm{x}.\ln \left(\frac{1}{\mathrm{x}}\right)}={\mathrm{e}}^{\underset{x\to 0}{\lim }-\tan \mathrm{x}.\ln \mathrm{x}}$$$$\mathrm{L}={\mathrm{e}}^{\underset{x\to 0}{\lim }\frac{-\ln \mathrm{x}}{\cot \mathrm{x}}}={\mathrm{e}}^{\underset{x\to 0}{\lim }\frac{-\frac{1}{\mathrm{x}}}{-\mathrm{cosec}{}^2\mathrm{x}}}$$$$\mathrm{L}={\mathrm{e}}^{\underset{x\to 0}{\lim }\frac{\sin {}^2\mathrm{x}}{\mathrm{x}}}={\mathrm{e}}^0=1$$

Answered by Dwaipayan Shikari last updated on 10/Sep/20

$$5)$$$$\underset{x\to 0}{\lim }\frac{e^x-e^{-x}}{sinx}=\frac{e^x-1}{x}+\frac{1-\frac{1}{e^x}}{x}=1+\frac{e^x-1}{e^{x}x}=2$$

Answered by Dwaipayan Shikari last updated on 10/Sep/20

$$\underset{x\to 0}{\lim }\frac{e^{x^2}-1}{1-cosx}=\underset{x\to 0}{\lim }\frac{e^{x^2}-1}{2{sin}^2\frac{x}{2}}=2\frac{e^{x^2}-1}{x^2}=2$$

Answered by mathmax by abdo last updated on 10/Sep/20

$$5)\mathrm{let}\mathrm{use}\mathrm{hospital}\mathrm{theorem}$$$${\lim }_{\mathrm{x}\to 0}\frac{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}}{\mathrm{sinx}}={\lim }_{\mathrm{x}\to 0}\frac{2\mathrm{sh}\left(\mathrm{x}\right)}{\mathrm{sinx}}={\lim }_{\mathrm{x}\to 0}\frac{2\mathrm{ch}\left(\mathrm{x}\right)}{\mathrm{cosx}}=2$$

Answered by mathmax by abdo last updated on 10/Sep/20

$$6)\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{e}}^{{\mathrm{x}}^2}-1}{1-\mathrm{cosx}}\mathrm{we}\mathrm{have}{\mathrm{e}}^{\mathrm{u}}\sim 1+\mathrm{u}\Rightarrow {\mathrm{e}}^{{\mathrm{x}}^2}\sim 1+{\mathrm{x}}^2\mathrm{and}$$$$1-\mathrm{cosx}\sim \frac{{\mathrm{x}}^2}{2}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\sim \frac{{\mathrm{x}}^2}{\frac{{\mathrm{x}}^2}{2}}=2\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=0$$

Answered by mathmax by abdo last updated on 10/Sep/20

$$7)\mathrm{let}\mathrm{g}\left(\mathrm{x}\right)={\left(\frac{1}{\mathrm{x}}\right)}^{\mathrm{tanx}}\Rightarrow \mathrm{g}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{tanxln}\left(\frac{1}{\mathrm{x}}\right)}={\mathrm{e}}^{-\ln \left(\mathrm{x}\right)\mathrm{tanx}}$$$$\Rightarrow \mathrm{g}\left(\mathrm{x}\right)\sim {\mathrm{e}}^{-\mathrm{xlnx}}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{g}\left(\mathrm{x}\right)={\mathrm{e}}^0=1$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com