Help with a Limit I have tried to solve this limit with L'Hopital's rule;
$\lim_{x\to0+}(1+x)^{lnx}$ but i applied several times with any result.
I would appreciate if somebody can help me.
Note: Wolfram alpha said that the result is 1, but it doesn't appear the steps to get it.
 A: hint: write $(1+x)^{\ln(x)}$ as $e^{\frac{\ln(1+x)}{\frac{1}{\ln(x)}}}$
A: To apply l'Hopital's rule, you first need to get the something in the form $\lim_{x\to a}\frac{f(x)}{g(x)}$, where $f(x)$ and $g(x)\to 0$ as $x\to a$. In you case, if we let $L=\lim_{x\to0^+}(1+x)\ln x$, then taking $\ln$ of both sides, we get
$$
\ln L = \lim_{x\to 0^+}\ln x\ln (1+x)=\lim_{x\to 0^+}\frac{\ln (1+x)}{1/\ln x}
$$
Now, you can apply l'Hopital's rule to the last expression, allowing you to find $\ln L$, then raise $e$ to that to find $L$.
A: Without using Hospital rule also we can evaluate this limit.
Substitute  $x=e^t$ then $x\to0^+$ is same as $t\to -\infty.$ $$\lim_{x\to0^+}(1+x)^{\ln x}=\lim_{t\to-\infty} (1+e^t)^t=\lim_{t\to-\infty} (1+O(e^t))=1.$$
A: Hint: $f(x,y)=x^y$ is continuous at $(e,0)$ so
$$
\begin{align}
\lim_{x\to0}\left[\left(1+x\right)^{1/x}\right]^{x\log(x)}
&=\left[\lim_{x\to0}\left(1+x\right)^{1/x}\right]^{\lim\limits_{x\to0}x\log(x)}\\
\end{align}
$$
You can also write $x\log(x)=\dfrac{\log(x)}{1/x}$ and use L'Hospital once.

Alternate Approach
$$
\begin{align}
\lim_{x\to0}\log\left(\left(1+x\right)^{\log(x)}\right)
=\lim_{x\to0}\frac{\log(x)}{1/x}\lim_{x\to0}\frac{\log(1+x)}{x}
\end{align}
$$
Each limit requires one application of L'Hospital.
