How can I evaluate the following limit $\lim_{x\to+ \infty} x^2\left(e^{\frac{1}{x}} -e^{\frac{1}{x+1}}\right)$? Evaluate this limit :
$$\lim_{x\to+ \infty} x^2\left(e^{\frac{1}{x}} -e^{\frac{1}{x+1}}\right)$$
I tried to simplify the function :
$$x^2e^{\frac{1}{x}}-x^2e^{\frac{1}{x+1}}=\frac{e^{\frac{1}{x}}}{\frac{1}{x^2}}-\frac{e^{\frac{1}{x+1}}}{\frac{1}{x^2}}$$
So by the substitution $X=\frac{1}x$  I'll have :
$$\lim_{X\to 0} \frac{e^X}{X^2} -\frac{e^{\frac{X}{X+1}}}{X^2}$$
Am I in the right path ?
 A: We have : \begin{aligned}\lim_{x\to +\infty}{x^{2}\left(\operatorname{e}^{\frac{1}{x}}-\operatorname{e}^{\frac{1}{x+1}}\right)}&=\lim_{x\to +\infty}{x^{2}\operatorname{e}^{\frac{1}{x+1}}\left(\operatorname{e}^{\frac{1}{x}-\frac{1}{x+1}}-1\right)}\\ &=\lim_{x\to +\infty}{\frac{x\operatorname{e}^{\frac{1}{x+1}}}{x+1}\times\frac{}{}\frac{\operatorname{e}^{\frac{1}{x\left(x+1\right)}}-1}{\frac{1}{x\left(x+1\right)}}}\\ &=1\times 1\\ \lim_{x\to +\infty}{x^{2}\left(\operatorname{e}^{\frac{1}{x}}-\operatorname{e}^{\frac{1}{x+1}}\right)}&=1\end{aligned}
Because $ \lim\limits_{x\to +\infty}{\frac{\operatorname{e}^{\frac{1}{x\left(x+1\right)}}-1}{\frac{1}{x\left(x+1\right)}}}=\lim\limits_{X\to 0}{\frac{\operatorname{e}^{X}-1}{X}}=1 $.
A: I think you're already on the right path, try to use the classical equivalents :
$$e^x\overset{x\to 0}\sim x+1$$
Using your results :
When $x$ tends to $0$ you'll have :
\begin{align}
\lim_{x\to 0} \frac{e^x}{x^2} -\frac{e^{\frac{x}{x+1}}}{x^2}&= \lim_{x\to 0} \frac{x+1}{x^2}-\frac{\frac{x}{x+1}+1}{x^2}\\
&=\lim_{x\to 0} \frac{x+1-\frac{x}{x+1}-1}{x^2}\\
&=\lim_{x\to 0} \frac{x^2+x-x}{x^2(x+1)}\\
&=\lim_{x\to 0} \frac{1}{x+1}\\
&=1\end{align}
Therefore your limit is $1$.
You may take this equivalence as a technique to evaluate limits :
$$e^{f(x)} \overset{f(x)\to 0}\sim f(x)+1$$
Make sure that you have :
$$f(x)\to 0$$
