1. ## Limit (no hospital)

Calculate :
$
\lim_{x\rightarrow +\infty }(e^{x}-1)^{\frac{1}{x}}
$

2. Substitute y = 1/x, and then evaluate:

$
\lim_ {y \to 0} (e ^ {1/y} -1) ^ y
$

You can then expand:

$
(e ^ {1/y} -1) ^ y = (e^ {1/y} )^y - y (e^ {1/y} )^{y-1} + \frac { y (y-1)} 2 (e^ {1/y} )^{y-2} - ...
$

$= e^1 - y e ^ {(\frac {y-1} y )} + \frac { y (y-1)} 2 (e^ {\frac {y-2} y )} - ...
$

You can see that as $y \to 0$, you're left with just e.

3. $\lim_{x\to\infty}(e^x-1)^{\frac{1}{x}}=\lim_{x\to\infty}e^{\displaystyle \ln(e^x-1)^{\frac{1}{x}}}=e^{\displaystyle\lim_{x\to\infty }\frac{\ln(e^x-1)}{x}}$

Now, $\frac{\ln(e^x-1)}{x}=\frac{\ln\left[e^x\left(1-\frac{1}{e^x}\right)\right]}{x}=\frac{x+\ln\left(1-\frac{1}{e^x}\right)}{x}=$

$=1+\frac{\ln\left(1-\frac{1}{e^x}\right)}{-\frac{1}{e^x}}\cdot\left(-\frac{1}{xe^x}\right)\to 1+1\cdot 0=1$

So the limit is $e$.

4. Thanks
I'have anthor solution :
$
\lim_{X\to+ \infty }(e^{x}-1)^{\frac{1}{x}}=\lim_{x\to +\infty }(e^{x}(1-e^{-x}))^{\frac{1}{x}}
$

$
\lim_{x\to +\infty }(e^{x})^{\frac{1}{x}}(1-e^{-x})^{\frac{1}{x}}=\lim_{x\to +\infty }e(1-e^{-x})^{\frac{1}{x}}=e\times 1=e
$