L'Hopital's Rule for nested exponential function

Question

I am trying to look for the following limit: $$\lim_{x\to\infty} (xe^{1/x} - x)^x$$

I re-wrote the limit in the following way

$$\exp\left( \lim_{x\to\infty} x\ln \left[ x(e^{1/x} -1) \right] \right)$$

and I look for the limit of the exponent using L'Hopital's rule (beforehand, I changed the $x$ term adjacent to the $\ln$ term so that it becomes $1/x$). However, I only end up with more complicated expressions as I repeatedly use L'Hopital's rule (I checked first if the resulting function after applying L'Hopital's is of the form $0/0$ or $\infty/\infty$). Is there more efficient way to solve this?

Answer 1

Hint

One of my old professors used to say (I quote him) :

"We shall always be closer to $0$ than to $\infty$".

So, obeying him,

$$\lim_{x\to\infty} (xe^{1/x} - x)^x=\lim_{t\to 0} \left(\frac{e^t-1}{t}\right)^{\frac{1}{t}}$$

Take logarithms, etc, etc ..;

Answer 2

I would use the Taylor series and expand the limit

$$ \lim_{x\to\infty} \left[x \left( 1 + \frac{1}{x} + \frac{1}{2x^2} \right) - x \right]^x = \lim_{x\to\infty} \left(1 + \frac{1}{2x} \right)^x = \sqrt e $$ by a definition of $e$.