Question Number 107028 by bemath last updated on 08/Aug/20
$$\:\:\:@{bemath}@ \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{2}+\mathrm{3}{x}\right)^{\frac{\mathrm{1}}{\mathrm{4}{x}−\mathrm{2}}} \\ $$
Commented by kaivan.ahmadi last updated on 08/Aug/20
$${y}=\left(\mathrm{2}+\mathrm{3}{x}\right)^{\frac{\mathrm{1}}{\mathrm{4}{x}−\mathrm{2}}} \Rightarrow{lny}=\frac{{ln}\left(\mathrm{2}+\mathrm{3}{x}\right)}{\mathrm{4}{x}−\mathrm{2}}\Rightarrow{lim}_{{x}\rightarrow\infty} {lny}= \\ $$$${lim}_{{x}\rightarrow\infty} \frac{{ln}\left(\mathrm{2}+\mathrm{3}{x}\right)}{\mathrm{4}{x}−\mathrm{2}}\sim{lim}_{{x}\rightarrow\infty} \frac{\mathrm{3}}{\mathrm{4}\left(\mathrm{2}+\mathrm{3}{x}\right)}=\mathrm{0} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\infty} {y}={e}^{\mathrm{0}} =\mathrm{1} \\ $$
Answered by Dwaipayan Shikari last updated on 08/Aug/20
$$\mathrm{y}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2}+\mathrm{3x}\right)^{\frac{\mathrm{1}}{\mathrm{4x}−\mathrm{2}}} \\ $$$$\mathrm{logy}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{log}\left(\mathrm{2}+\mathrm{3x}\right)}{\mathrm{4x}−\mathrm{2}}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}}{\mathrm{4}\left(\mathrm{2}+\mathrm{3x}\right)}=\mathrm{0} \\ $$$$\mathrm{y}=\mathrm{e}^{\mathrm{0}} =\mathrm{1} \\ $$
Answered by john santu last updated on 08/Aug/20
![@JS@ lim_(x→∞) (2+3x)^(1/(4x−2)) =lim_(x→∞) (1+(1/(((1/(3x+1))))))^(1/(4x−2)) = [lim_(x→∞) (1+(1/(((1/(3x+1))))))^(3x+1) ]^(1/((3x+1)(4x−2))) = e^(lim_(x→∞) ((1/((3x+1)(4x−2))))) = e^0 =1](https://www.tinkutara.com/question/Q107060.png)
$$\:\:\:\:\:@\mathrm{JS}@ \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2}+\mathrm{3x}\right)^{\frac{\mathrm{1}}{\mathrm{4x}−\mathrm{2}}} \:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{3x}+\mathrm{1}}\right)}\right)^{\frac{\mathrm{1}}{\mathrm{4x}−\mathrm{2}}} \\ $$$$=\:\left[\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{3x}+\mathrm{1}}\right)}\right)^{\mathrm{3x}+\mathrm{1}} \right]^{\frac{\mathrm{1}}{\left(\mathrm{3x}+\mathrm{1}\right)\left(\mathrm{4x}−\mathrm{2}\right)}} \\ $$$$=\:\mathrm{e}\:^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\left(\mathrm{3x}+\mathrm{1}\right)\left(\mathrm{4x}−\mathrm{2}\right)}\right)} =\:\mathrm{e}^{\mathrm{0}} =\mathrm{1} \\ $$
Answered by mathocean1 last updated on 08/Aug/20
$$=>\:{let}\:\mathrm{2}+\mathrm{3}{x}={a}. \\ $$$$\underset{{x}\rightarrow\infty\:} {{lim}}\:{a}^{\frac{\mathrm{1}}{\mathrm{4}\left(\infty\right)−\mathrm{2}}} =\underset{{x}\rightarrow\infty} {{lim}}\:{a}^{\mathrm{0}} =\underset{{x}\rightarrow\infty} {{lim}}\:\mathrm{1}=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$“{ocean}'' \\ $$