Find the limit $\lim_{x\to\infty} \left(e^{\frac{1}{x}}+\frac{1}{x}\right)^{x}$ $\lim_{x\to\infty}  \left(e^{\frac{1}{x}}+\dfrac{1}{x}\right)^{x}$
Can you help me to find this limit. I am unable to find a way to finish this beast. Even though I have done some similar limits involving $e$ number.
Should I take the logarithm and use L'Hopital rule?
 A: simply thinking
$$e^x = 1 + x + O(x^2) \quad (x\to0)$$
so
$$\left(e^{\frac1{x}}+\frac1{x}\right)^x =\left(1+\frac2{x}+O(x^{-2})\right)^x\to e^2 \quad (x\to\infty)$$
edit for some supplement, in general
if you meet $(1+x+o(x))^{\frac1{x}}$ type limitation for $x\to0$
you always have
$$\frac1{x}\ln(1+x+o(x))=\frac1{x}(x+o(x)+o(x+o(x)))\to1$$
hence you will produce $e$ in result. in your case, you write
$$\left(1+\frac2{x}+O(x^{-2})\right)^x=\left(1+\frac2{x}+O(x^{-2})\right)^{\frac{x}{2}\cdot2}$$ leads to $e^2$
moreover $(1+f(x)+o(f(x)))^{\frac1{g(x)}}$ always leads to $e^z$ and $z$ is almostly decided by the limitation of $\frac{f(x)}{g(x)}$ (not that strict, but you always have $\frac1{g(x)}=\frac1{f(x)}\frac{f(x)}{g(x)}$ on shoulder), so you may deal with it by this fashion, what i write before is a rough thinking to get the result.
A: Considering
$$f(x)=\left(e^{1/x}+\frac1x\right)^x$$
we have that
$$\begin{align}
\lim_{x\to\infty}\ln{(f(x))}
&=\lim_{x\to\infty}x\ln{\left(e^{1/x}+\frac1x\right)}\\
&=\lim_{x\to\infty}x\left(\ln{\left(e^{1/x}\right)}+\ln{\left(1+\frac1{xe^{1/x}}\right)}\right)\\
&=\lim_{x\to\infty}\left(1+x\ln{\left(1+\frac1{xe^{1/x}}\right)}\right)\\
&=\lim_{x\to\infty}\left(1+x\left(\frac1{xe^{1/x}}+o\left(\frac1{xe^{1/x}}\right)\right)\right)\\
&=\lim_{x\to\infty}\left(1+\frac1{e^{1/x}}+o\left(\frac1{e^{1/x}}\right)\right)\\
&=2\\
\end{align}$$
Hence the limit is given by
$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}e^{\ln{(f(x))}}=e^{\lim_{x\to\infty}\ln{(f(x))}}=e^2$$
A: Replace $x$ by ${1 \over x}$. Your limit becomes
$$\lim_{x \rightarrow 0^+} (e^x + x)^{1 \over x}$$
Taking logarithms, the natural log of your limit is
$$\lim_{x \rightarrow 0^+} {\ln(e^x + x) \over x}$$
Now you can use l'hopital's rule to get that this is the same as
$$\lim_{x \rightarrow 0^+} {e^x + 1 \over e^x + x}$$
Plug in $0$ to get the limit being $2$. This is the natural log of the original limit, so your original limit is $e^2$.
A: I wish to try this:
$\lim_{x\to\infty}  \left(e^{\frac{1}{x}}+\dfrac{1}{x}\right)^{x}= \lim_{x\to\infty}  e^{\ln\left(e{\frac{1}{x}}+\dfrac{1}{x}\right)^{x}}= \lim_{x\to\infty}e^{x\ln\left(e^{\frac{1}{x}}+\dfrac{1}{x}\right)}=\lim_{x\to\infty}e^\dfrac{{\ln\left(e^{\frac{1}{x}}+\dfrac{1}{x}\right)}}{\dfrac{1}{x}}=\lim_{x\to\infty}e{\dfrac{\dfrac{d}{dx}\ln\left(e^{\frac{1}{x}}+\dfrac{1}{x}\right)}{\dfrac{d}{dx}(\dfrac{1}{x})}}=\lim_{x\to\infty}e^\left(\dfrac{\dfrac{-e^{\frac{1}{x}}-1}{x(xe^{\frac{1}{x}}+1)}}{-\dfrac{1}{x^2}}\right)=\lim_{x\to\infty}e^\left(\dfrac{-x^{2}(-e^{\frac{1}{x}}-1)}{x(xe^{\frac{1}{x}}+1)}\right)=\lim_{x\to\infty}e^\left(\dfrac{-x(-e^{\frac{1}{x}}-1)}{xe^{\frac{1}{x}}+1}\right)=\lim_{x\to\infty}e^\left(\dfrac{\dfrac{-x(-e^{\frac{1}{x}}-1)}{x}}{\dfrac{xe^{\frac{1}{x}}+1}{x}}\right)=\lim_{x\to\infty}e^\left(\dfrac{e^{\frac{1}{x}}+1)}{{e^{\frac{1}{x}}+\dfrac{1}{x}}}\right)=e^{\dfrac{1+1}{1+0}}=e^2$
The answer seems to be $e^{2}$, so I must make some mistakes along the way. This is the most challenging post I have ever posted using latex.
Edit: THanks to A.Γ., I have been able to correct all steps.
A: Letting $u=1/x\to0^+$ and using the limit theorem $\lim(f^g)=(\lim f)^{\lim g}$ if $\lim f$ and $\lim g$ both exist and aren't both equal to $0$ (and $f$ avoids negative values, so that $f^g$ is well defined), we have
$$\left(e^{1/x}+{1\over x} \right)^x=(e^u+u)^{1/u}=e\left(\left(1+{u\over e^u}\right)^{e^u/u}\right)^{1/e^u}\to e(e)^1=e^2$$
since $(1+1/w)^w\to e$ as $w=e^u/u\to\infty$.
A: Let $u=\frac1x$ and then
\begin{eqnarray}
&&\lim_{x\to\infty}x\ln(e^{\frac1x}+\frac1x)\\
&=&\lim_{u\to0}\frac{\ln(e^u+u)}{u}=\lim_{u\to0}\frac{\ln[1-(1-e^u-u)]}{u}\\
&=&\lim_{u\to0}\frac{\ln[1-(1-e^u-u)]}{1-e^u-u}\cdot\frac{1-e^u-u}{u}\\
&=&(-1)(-2)=2
\end{eqnarray}
and hence
$$ \lim_{x\to\infty}(e^{\frac1x}+\frac1x)^x=e^2.$$
Here
$$ \lim_{x\to0}\frac{\ln(1-x)}{x}=\lim_{x\to0}\frac{1-e^x}{x}=-1$$
are used.
