Solving finite limit without L'Hôpital I have come across a problem which requires solving the following limit without L'Hôpital rule:
$$\lim_{x\to\infty} x^2\cdot(e^\frac{1}{x-1}-e^\frac{1}{x})$$
It is obvious from the graphic plot (or using L'Hôpital rule) that the limit is 1. I have tried a few algebraic manipulations and changes of variable without success. Another approach that I tried was to "sandwich" this limit between two other different limits, one strictly greater and one strictly lesser than this one, that would be easier to work with than the difference of exponentials present in this one. As of now I haven't had any success, how would one go about solving it?

By the mean value theorem,
$$e^{\frac{1}{x-1}}-e^{\frac{1}{x}} = e^{c_x}\left(\frac{1}{x-1} -\frac{1}{x}\right
)=e^{c_x}\left(\frac{1}{(x-1)x}\right
)$$
for some $c_x$ between $1/x$ and $1/(x-1).$ Thus as $x\to \infty, c_x\to 0,$ hence $e^{c_x}\to 1.$ Multiplying by $x^2$ then gives a limit of $1.$ 
