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Question Number 107030 by bemath last updated on 08/Aug/20

$$\underset{x\to 0}{\lim }{(2+\frac{3}{x})}^{\frac{1}{4x-2}}$$

Commented by kaivan.ahmadi last updated on 08/Aug/20

$$y={(2+\frac{3}{x})}^{\frac{1}{4x-2}}\Rightarrow {lim}_{x\to \mathrm{\infty }}lny={lim}_{x\to \mathrm{\infty }}\frac{ln(2+\frac{3}{x})}{4x-2}\sim$$$${lim}_{x\to \mathrm{\infty }}\frac{-3}{4x^2(2+\frac{3}{x})}=0\Rightarrow {lim}_{x\to \mathrm{\infty }}y=e^0=1$$

Answered by Dwaipayan Shikari last updated on 08/Aug/20

$$\mathrm{y}=\underset{x\to 0}{\lim }{(2+\frac{3}{\mathrm{x}})}^{\frac{1}{4\mathrm{x}-2}}$$$$\mathrm{logy}=\underset{x\to 0}{\lim log}(2+\frac{3}{\mathrm{x}}).\frac{1}{4\mathrm{x}-2}$$$$\mathrm{logy}=\underset{x\to 0}{\lim }(-\frac{1}{2}\log (2+\frac{3}{\mathrm{x}}))$$$$\mathrm{logy}\to -\mathrm{\infty }$$$$\mathrm{y}=0$$

Commented by bemath last updated on 08/Aug/20

$$notcorrectsir$$

Commented by Dwaipayan Shikari last updated on 08/Aug/20

Commented by bemath last updated on 08/Aug/20

Answered by john santu last updated on 08/Aug/20

$$@\mathrm{JS}@$$$$\mathrm{we}\mathrm{have}\mathrm{formula}\underset{x\to 0}{\lim }{(1+\mathrm{x})}^{\frac{1}{\mathrm{x}}}=\mathrm{e}$$$$\underset{x\to 0}{\lim }{(1+\left(\frac{\mathrm{x}+3}{\mathrm{x}}\right))}^{\frac{1}{4\mathrm{x}-2}}=\underset{x\to 0}{\lim }{\left[{(1+\left(\frac{\mathrm{x}+3}{\mathrm{x}}\right))}^{\left(\frac{\mathrm{x}}{\mathrm{x}+3}\right)}\right]}^{\frac{\mathrm{x}+3}{\mathrm{x}(4\mathrm{x}-2)}}$$$$=\mathrm{e}{}^{\underset{x\to 0}{\lim }\left(\frac{\mathrm{x}+3}{{4\mathrm{x}}^2-2\mathrm{x}}\right)}={\mathrm{e}}^{\mathrm{\infty }}=\mathrm{\infty }$$

Commented by bemath last updated on 08/Aug/20

$$thankyou$$

Answered by bemath last updated on 08/Aug/20

$$\underset{x\to 0}{\lim }{\left(\frac{2x+3}{x}\right)}^{\frac{1}{4x-2}}=y$$$$\ln y=\underset{x\to 0}{\lim }\frac{1}{4x-2}.\ln \left(\frac{2x+3}{x}\right)$$$$\ln y=\underset{x\to 0}{\lim }\frac{\ln \left(\frac{2x+3}{x}\right)}{4x-2}$$$$\ln y=\underset{x\to 0}{\lim }\frac{\frac{d}{dx}\left(\ln \left(\frac{2x+3}{x}\right)\right)}{\frac{d}{dx}(4x-2)}$$$$\ln y=\underset{x\to 0}{\lim }\frac{\left(\frac{x}{2x+3}\right)(-\frac{3}{x^2})}{4}=\mathrm{\infty }$$$$thereforey=e^{\mathrm{\infty }}=\mathrm{\infty }$$

Commented by Dwaipayan Shikari last updated on 08/Aug/20

$$\mathrm{It}\mathrm{reaches}\mathrm{to}{\mathrm{e}}^{\mathrm{p}}=0(\mathrm{p}\to -\mathrm{\infty })$$

Answered by mathmax by abdo last updated on 08/Aug/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={(2+\frac{3}{\mathrm{x}})}^{\frac{1}{4\mathrm{x}-2}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\frac{1}{4\mathrm{x}-2}\ln (2+\frac{3}{\mathrm{x}})}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\frac{1}{\mathrm{x}(4-\frac{2}{\mathrm{x}})}\ln (2+\frac{3}{\mathrm{x}})}\mathrm{we}\mathrm{do}\mathrm{tbe}\mathrm{changement}\frac{1}{\mathrm{x}}=\mathrm{t}$$$$(\mathrm{x}\to 0^{+}\Rightarrow \mathrm{t}\to +\mathrm{\infty })\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\frac{\mathrm{t}}{4-2\mathrm{t}}\ln (2+3\mathrm{t})}\mathrm{but}$$$${\lim }_{\mathrm{t}\to +\mathrm{\infty }}\frac{\mathrm{t}}{4-2\mathrm{t}}\ln (3\mathrm{t}+2)={\lim }_{\mathrm{t}\to +\mathrm{\infty }}-\frac{1}{2}\ln (3\mathrm{t}+2)=-\mathrm{\infty }\Rightarrow$$$${\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=0$$

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