Find $\lim_{x\to\infty}x^2 \left( e^{1/x}-e^{1/(x+1)} \right)$ without using the L'Hopital's rule or Taylor's series

Question

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.

Answer 1

$$\lim_{x\to\infty}x^2 \left( e^{\frac1x}-e^{\frac{1}{x+1}} \right)=\lim_{x\to\infty}x^2 e^{\frac{1}{x+1}}\left( e^{\frac1x-\frac{1}{x+1}}-1 \right)=\\ =\lim_{x\to\infty}x^2 \left( e^{\frac{1}{x(x+1)}}-1 \right)=\lim_{x\to\infty}x^2\frac{1}{x(x+1)} = 1$$ 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

Answer 2

$$ L=\lim_{x\to\infty}x^2 \left( e^{1/x}-e^{1/(x+1)} \right). $$ Let $t=1/x$, $\frac{1}{1+x}=t(1+t)^{-1}=t(1-t+t^2+O(t^3))$, then $$L=\lim_{t \to 0} \frac{e^t-e^{t-t^2+O(t^3))}}{t^2}=\lim_{t\to 0} e^{t}\frac{1-e^{-t^2}}{t^2}=\lim_{t\to 0}e^t \frac{1-1+t^2+O(t^4)}{t^2}=1.$$