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

Question

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)

Answer 1

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

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

Answer 2

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.

Answer 3

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

Answer 4

$$ \lim _{x\to \infty }\left(\left(e^x-1\right)^{\frac{1}{x}}\right)=\lim_{x\rightarrow \infty}e^{\frac{1}{x}\ln(e^x-1)}\approx \lim_{x\rightarrow \infty}\frac{1}{x}\ln(e^x)=\lim_{x\rightarrow \infty}\frac{x}{x}=1 \rightarrow \color{red}{e^1} $$

Answer 5

$$ \begin{gathered} \mathop {\lim }\limits_{x\; \to \;\infty } \left( {e^{\,x} - 1} \right)^{\,1/x} = \mathop {\lim }\limits_{x\; \to \;\infty } e^{\,x/x} \left( {1 - e^{\, - \,x} } \right)^{\,1/x} = \hfill \\ = e\mathop {\lim }\limits_{x\; \to \;\infty } \left( {1 - e^{\, - \,x} } \right)^{\,1/x} = e\mathop {\lim }\limits_{y\; \to \;0} \left( {1 - e^{\, - \,\frac{1} {y}} } \right)^{\,y} = \hfill \\ = e\mathop {\lim }\limits_{y\; \to \;0} \left( {\left( \begin{gathered} y \\ 0 \\ \end{gathered} \right)e^{\, - \,\frac{0} {y}} - \left( \begin{gathered} y \\ 1 \\ \end{gathered} \right)e^{\, - \,\frac{1} {y}} + \left( \begin{gathered} y \\ 2 \\ \end{gathered} \right)e^{\, - \,\frac{2} {y}} + \cdots } \right) = \hfill \\ = e \hfill \\ \end{gathered} $$

Answer 6

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.

Answer 7

$$\lim\limits_{x\rightarrow+\infty}\ln(e^x-1)^{\frac{1}{x}}=\lim\limits_{x\rightarrow+\infty}\frac{\ln(e^x-1)}{x}=\lim\limits_{x\rightarrow+\infty}\frac{\ln\left(1-\frac{1}{e^x}\right)e^x}{x}=$$ $$=\lim\limits_{x\rightarrow+\infty}\frac{x+\ln\left(1-\frac{1}{e^x}\right)}{x}=1+\lim\limits_{x\rightarrow+\infty}\frac{\ln\left(1-\frac{1}{e^x}\right)}{x}=1+0=1,$$ which says that the answer is $e$.