Solve limit with Lagrange theorem I tried to solve this limit:
$$ \lim_{x \to +\infty} x^2\left(e^{\frac{1}{x+1}}-e^{\frac{1}{x}}\right) $$
Instead of solving it with Taylor series (using $u = 1/x$), I noticed that the difference within the parenthesis is the $\Delta f$ of the function $e^{\frac{1}{x}}$.

$\lim\limits_{x \to +\infty} x^2 * \Delta\left(e^{1/x}\right) = $
The Lagrange theorem says that $\Delta f = D\left[f\right] * \Delta x$, so
$=\lim\limits_{x \to +\infty} x^2 * \left( D\left[e^{1/x}\right]_{x_0} * \Delta x\right), x_0 \in \left(x, x+1\right)$
$$ x_0=x+r(x)$$ 
  $$ 0<r(x)<1 $$
$=\lim\limits_{x \to +\infty} x^2 * \left( e^{\frac{1}{x+r(x)}} * \left(-\frac{1}{\left(x+r(x)\right)^2}\right)\right) $
$=\lim\limits_{x \to +\infty} -\frac{1}{\left(x+r(x)\right)^2} * x^2 * e^{\frac{1}{x+r(x)}}$
$\approx \lim\limits_{x \to +\infty} -\frac{1}{x^2} * x^2 * e^{\frac{1}{x}} $
$=\lim\limits_{x \to +\infty} -1 * e^{\frac{1}{x}} = -1$

Is it correct to proceed in this way?
EDIT: Using approximations is not a very elegant solution. I might be more precise using the Squeeze theorem (I'll think the solution, then I'll post it)
Instead of using the approximation, we consider these three functions:


*

*$f(x) = x^2 * e^{\frac{1}{x+1}} * \frac{1}{(x+1)^2}$

*$h(x) = x^2 * e^{\frac{1}{x+r(x)}} * \frac{1}{(x+r(x))^2}$

*$g(x) = x^2 * e^{\frac{1}{x}} * \frac{1}{x^2}$


Obviously, $f(x) \le h(x) \le g(x)$ for $x > 0$ and $r(x) \in (0, 1)$.

$\lim\limits_{x \to +\infty} f(x) = \lim\limits_{x \to +\infty} x^2 * e^{\frac{1}{x+1}} * \frac{1}{(x+1)^2} = 1$
(It is 1 and not -1 just because I left the minus sign out of the limit)
$\lim\limits_{x \to +\infty} g(x) = \lim\limits_{x \to +\infty} x^2 * e^{\frac{1}{x}} * \frac{1}{x^2} = 1$

So $\lim\limits_{x \to +\infty} h(x) = 1 \rightarrow \lim\limits_{x \to +\infty} -h(x) = \lim\limits_{x \to +\infty} x^2\left(e^{\frac{1}{x+1}}-e^{\frac{1}{x}}\right) = -1$
Here's how the f (the red function) and g (the blue function) look like:


\begin{align}
& \lim_{x \rightarrow \infty} x^2 \left( e^{\frac{1}{x+1}} -e^{\frac{1}{x}} \right)\\
= & \lim_{x \rightarrow \infty} \left( 1+ \frac{1}{x+1} + \frac{1}{2 (x+1)^2} + O(x^{-3}) - 1 -\frac{1}{x} - \frac{1}{2 x^2} + O(x^{-3})\right) \\
= & \lim_{x \rightarrow \infty} \frac{-x^2}{(x+1)x} + \frac{-2x-1}{2(x+1)^2 x^2} +O(x^{-1})
= -1.
\end{align}
