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Question Number 107028 by bemath last updated on 08/Aug/20

$$@bemath@$$$$\underset{x\to \mathrm{\infty }}{\lim }{(2+3x)}^{\frac{1}{4x-2}}$$

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

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

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

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

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

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

Answered by mathocean1 last updated on 08/Aug/20

$$=>let2+3x=a.$$$$\underset{x\to \mathrm{\infty }}{lim}{a}^{\frac{1}{4\left(\mathrm{\infty }\right)-2}}=\underset{x\to \mathrm{\infty }}{lim}{a}^0=\underset{x\to \mathrm{\infty }}{lim}1=1$$$$$$$$$$$$‘‘{ocean}^{″}$$

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