limit of $(e^x-1)^{\frac{1}{x}}$ - Without L'hopital I am having trouble calculating the following limit:
$$\lim_{x\to\infty}(e^x-1)^{\frac{1}{x}}$$
I figured out (With help from wolfram) that the limit is $e$, but I can't understand why.
I tried to use $x^a = e^{a\ln(x)}$ but that didn't help because the base is $e^x-1$
Any hint or explanation is welcome!
PS: I can't use L'hopital (tagged so specifically)
 A: Note that we can write
$$\left(e^x-1\right)^{1/x}=e\left(1-e^{-x}\right)^{1/x}$$
The limit, $\lim_{x\to \infty}\left(1-e^{-x}\right)^{1/x}$, is not of indeterminate form since $1^0=1$.
Therefore, we have
$$\begin{align}
\lim_{x\to \infty}\left(e^x-1\right)^{1/x}&=\lim_{x\to \infty}\left(e\left(1-e^{-x}\right)^{1/x}\right)\\\\
&=e\lim_{x\to \infty}\left(1-e^{-x}\right)^{1/x}\\\\
&=e
\end{align}$$ 
A: Check that $(1/2)e^x < e^x-1$ for $x>1.$ Thus
$$((1/2)e^x)^{1/x} < (e^x-1)^{1/x}< (e^x)^{1/x}\,\text { for } x>1.$$
The term on the left equals $(1/2)^{1/x}e,$ the term on the right equals $e.$ Since $(1/2)^{1/x} \to 1,$ the desired limit is $e$ by the squeeze theorem.
A: If 
$$
L=\lim_{x\rightarrow \infty}(e^x-1)^{1/x}\Rightarrow \log L=\log\lim_{x\rightarrow \infty}(e^x-1)^{1/x}\stackrel{\text{continuity of the logarithm}}{\Rightarrow} \lim_{x\rightarrow \infty}1/x\log(e^x-1)\stackrel{L'Hôpital's rule}=\lim_{x\rightarrow \infty}\frac{e^x}{e^x}=1\Rightarrow\log L=1\Rightarrow e=L
$$
To remove L'Hôpital's use:
$$
\lim_{x\rightarrow \infty}1/x\log(e^x-1)\sim \lim_{x\rightarrow \infty}1/x\log(e^x)=\lim_{x\rightarrow \infty}x/x=1
$$
A: We have: $\ln \left(\left(e^x-1\right)^{1/x}\right) = \dfrac{\ln(e^x-1)}{x}$. You can use L'hospital rule here..and your answer would be $e^L$ whereas $L$ is the limit of the L'hospital.
A: $$
\lim _{x\to \infty }\left(\left(e^x-1\right)^{\frac{1}{x}}\right)=\lim_{x\rightarrow \infty}e^{\frac{1}{x}\ln(e^x-1)}\approx \lim_{x\rightarrow \infty}\frac{1}{x}\ln(e^x)=\lim_{x\rightarrow \infty}\frac{x}{x}=1 \rightarrow \color{red}{e^1}
$$
A: $$
\begin{gathered}
  \mathop {\lim }\limits_{x\; \to \;\infty } \left( {e^{\,x}  - 1} \right)^{\,1/x}  = \mathop {\lim }\limits_{x\; \to \;\infty } e^{\,x/x} \left( {1 - e^{\, - \,x} } \right)^{\,1/x}  =  \hfill \\
   = e\mathop {\lim }\limits_{x\; \to \;\infty } \left( {1 - e^{\, - \,x} } \right)^{\,1/x}  = e\mathop {\lim }\limits_{y\; \to \;0} \left( {1 - e^{\, - \,\frac{1}
{y}} } \right)^{\,y}  =  \hfill \\
   = e\mathop {\lim }\limits_{y\; \to \;0} \left( {\left( \begin{gathered}
  y \\ 
  0 \\ 
\end{gathered}  \right)e^{\, - \,\frac{0}
{y}}  - \left( \begin{gathered}
  y \\ 
  1 \\ 
\end{gathered}  \right)e^{\, - \,\frac{1}
{y}}  + \left( \begin{gathered}
  y \\ 
  2 \\ 
\end{gathered}  \right)e^{\, - \,\frac{2}
{y}}  +  \cdots } \right) =  \hfill \\
   = e \hfill \\ 
\end{gathered} 
$$
A: $$\lim\limits_{x\rightarrow+\infty}\ln(e^x-1)^{\frac{1}{x}}=\lim\limits_{x\rightarrow+\infty}\frac{\ln(e^x-1)}{x}=\lim\limits_{x\rightarrow+\infty}\frac{\ln\left(1-\frac{1}{e^x}\right)e^x}{x}=$$
$$=\lim\limits_{x\rightarrow+\infty}\frac{x+\ln\left(1-\frac{1}{e^x}\right)}{x}=1+\lim\limits_{x\rightarrow+\infty}\frac{\ln\left(1-\frac{1}{e^x}\right)}{x}=1+0=1,$$
which says that the  answer is $e$.
