Question

$\lim_{x \to \infty} ({\frac{2e^3^x}{e^2^x+1}})^ \frac{1}{x}$

Answer 1

$\underset{x\rightarrow \infty}{lim} \Big(\dfrac{2e^{3x}}{e^{2x}+1} \Big)^ \dfrac{1}{x}}=\underset{x\rightarrow \infty}{lim} \dfrac{(2e^{3x})^ \dfrac{1}{x} }{(e^{2x}+1)^{\dfrac{1}{x}}} =\underset{x\rightarrow \infty}{lim} \dfrac{2^{\dfrac{1}{x}}\cdot e^{\dfrac{3x}{x}}}{(e^{2x}+1)^{\dfrac{1}{x}}} =$

$=\underset{x\rightarrow \infty}{lim} \dfrac{2^{\dfrac{1}{x}}\cdot e^{3}}{(e^{2x}+1)^{\dfrac{1}{x}}} =e^3\cdot \underset{x\rightarrow \infty}{lim} \dfrac{2^{\dfrac{1}{x}}}{(e^{2x}+1)^{\dfrac{1}{x}}} =e^3\cdot \dfrac{\underset{x\rightarrow \infty}{lim}2^{\dfrac{1}{x}}}{\underset{x\rightarrow \infty}{lim}(e^{2x}+1)^{\dfrac{1}{x}}} =$

$=e^3\cdot \dfrac{1}{\underset{x\rightarrow \infty}{lim}(e^{2x}+1)^{\dfrac{1}{x}}} \overset{(*)}{=} \\ \\ \underset{x\rightarrow \infty}{lim}(e^{2x}+1)^{\dfrac{1}{x}}= \underset{x\rightarrow \infty}{lim}e^\big{ln(e^{2x}+1)^{\dfrac{1}{x}}} =\underset{x\rightarrow \infty}{lim}e^\big{\dfrac{1}{x}\cdot ln(e^{2x}+1)} =$

$=\underset{x\rightarrow \infty}{lim}e^\big{ \dfrac{ln(e^{2x}+1)}{x}} = e^\big{\underset{x\rightarrow \infty}{lim} \dfrac{ln(e^{2x}+1)}{x}} \\ \\ \underset{x\rightarrow \infty}{lim} \dfrac{ln(e^{2x}+1)}{x} \overset{\frac{\infty}{\infty}(L'H.)}{=}\underset{x\rightarrow \infty}{lim} \dfrac{\dfrac{(e^{2x}+1)'}{e^{2x}+1}}{1}=\underset{x\rightarrow \infty}{lim} \dfrac{2e^{2x}}{e^{2x}+1} \overset{\frac{\infty}{\infty}(L'H.)}{=}$

$\overset{\frac{\infty}{\infty}(L'H.)}{=} \underset{x\rightarrow \infty}{lim} \dfrac{2\cdot2\cdot e^{2x}}{2\cdot e^{2x}}=2 \\ \\ \Rightarrow \boxed{e^\big{\underset{x\rightarrow \infty}{lim} \dfrac{ln(e^{2x}+1)}{x}}= e^2 \Leftrightarrow \underset{x\rightarrow \infty}{lim}(e^{2x}+1)^{\dfrac{1}{x}}=e^2} \\ \\ \\ \Rightarrow e^3\cdot \dfrac{1}{\underset{x\rightarrow \infty}{lim}(e^{2x}+1)^{\dfrac{1}{x}}} \overset{(*)}{=} e^3\cdot \dfrac{1}{e^2} = e$


$\Rightarrow\boxed{\boxed{\underset{x\rightarrow \infty}{lim} \Big(\dfrac{2e^{3x}}{e^{2x}+1} \Big)^ {\dfrac{1}{x}}=e}}}}}}$