This limit is proposed to be solved without using the L'Hopital's rule or Taylor series:
$$
\lim_{x\to\infty}x^2 \left( e^{1/x}-e^{1/(x+1)} \right).
$$
I know how to calculate this limit using the Taylor expansion:
$$
x^2 \left( e^{1/x}-e^{1/(x+1)} \right)=
x^2 \left( 1+\frac1{x}+\frac1{2!}\frac1{x^2}+\ldots – 1-\frac1{x+1}-\frac1{2!}\frac1{(x+1)^2}+\ldots \right)
$$
$$
=x^2 \left( \frac1{x(x+1)}+\frac1{2!}\frac{2x+1}{x^2(x+1)^2}+\ldots \right)=
\frac 1{1+\frac1{x}}+\frac1{2!}\frac{2x+1}{(x+1)^2}+\ldots,
$$
thus, the limit is equal to $1$.
I'm allowed to use the fact that $\lim_{x\to0}\frac{e^x-1}{x}=1$, but I don't know how to apply it to this problem.
Best Answer
$$\lim_{x\to\infty}x^2 \left( e^{\frac1x}-e^{\frac{1}{x+1}} \right)=\lim_{x\to\infty}x^2 e^{\frac{1}{x+1}}\left( e^{\frac1x-\frac{1}{x+1}}-1 \right)=\\ =\lim_{x\to\infty}x^2 \left( e^{\frac{1}{x(x+1)}}-1 \right)=\lim_{x\to\infty}x^2\frac{1}{x(x+1)} = 1$$ On last step we use $\lim_\limits{y\to0}\frac{e^y-1}{y}=1$ considering it with composition $y=\frac{1}{x(x+1)}$, knowing $\lim_\limits{x\to\infty}\frac{1}{x(x+1)}=0$. Exact conditions for composition/substitution you can find on my answer from Bringing limits inside functions when limits go to infinity