limit (e^x + e^(2x))^(2/x) as x tends to infinity

[turkey]

lim (e^x + e^(2x) ) ^(2/x)
x->infinity

 

[galactus]

\(\displaystyle \L\\\lim_{x\to\infty}(e^{x}+e^{2x})^{\frac{2}{x}}\)

Rewrite:

\(\displaystyle \L\\\lim_{x\to\infty}e^{\frac{2}{x}ln(e^{x}+e^{2x})}\)

\(\displaystyle \L\\e^{\lim_{x\to\infty}(\frac{2}{x}ln(e^{x}+e^{2x}))}\)

\(\displaystyle \L\\e^{2\lim_{x\to\infty}\frac{ln(e^{x}+e^{2x})}{x}}\)

Now, try L'Hopital's rule:

\(\displaystyle \L\\ln(e^{x}+e^{2x})dx=2-\frac{1}{e^{x}+1}\)

\(\displaystyle \L\\e^{2\lim_{x\to\infty}\frac{2-\frac{1}{e^{x}+1}}{1}}\)

As can be seen, the limit of \(\displaystyle \L\\2-\frac{1}{e^{x}+1}\rightarrow\2\)

So, we have:

\(\displaystyle \L\\e^{2(2)}=e^{4}\)

 

[pka]

Let’s do it without L'Hopital's rule!
Note that:
\(\displaystyle \L
\begin{array}{rcl}
e^4 & = & \left( {e^{2x} } \right)^{\left( {2/x} \right)} \\
& \le & \left( {e^x + e^{2x} } \right)^{\left( {2/x} \right)} \\
& \le & \left( {e^{2x} + e^{2x} } \right)^{\left( {2/x} \right)} \\
& = & \left( 2 \right)^{\left( {2/x} \right)} \left( {e^{2x} } \right)^{\left( {2/x} \right)} \\
& = & \left( 2 \right)^{\left( {2/x} \right)} \left( {e^4 } \right) \\
\end{array}\)

Now we know that \(\displaystyle \L
\lim _{n \to \infty } \left( 2 \right)^{\left( {2/x} \right)} = 1\).

So by the ‘squeeze play principle’ what is the limit?

 

[galactus]

Nice, pka.