limit of $(e^x-1)^{\frac{1}{x}}$ - Without L'hopital

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

$$\begin{align} \lim_{x\to \infty}\left(e^x-1\right)^{1/x}&=\lim_{x\to \infty}\left(e\left(1-e^{-x}\right)^{1/x}\right)\\\\ &=e\lim_{x\to \infty}\left(1-e^{-x}\right)^{1/x}\\\\ &=e \end{align}$$

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.

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: $$ \lim_{x\rightarrow \infty}1/x\log(e^x-1)\sim \lim_{x\rightarrow \infty}1/x\log(e^x)=\lim_{x\rightarrow \infty}x/x=1 $$