The gap between two exponentials at infinity

Idea. The two exponentials are very close to each other, since $\frac1x$ and $\frac1{x+1}$ are close. The gap is of order $\frac1{x^2}$, which $x^2$ compensates.

Computation. With $u=\frac1x$, $v=\frac1{x+1}$ and $w=u-v=\dfrac1{x(x+1)}>0$,
$$e^u-e^v=e^v\left(e^{w}-1\right).$$
Then
$$x^2\left(e^u-e^v\right)=e^v\cdot\frac{e^w-1}{w}\cdot x^2w=e^v\cdot\frac{e^w-1}{w}\cdot\frac{x}{x+1}.$$
As $x\to+\infty$: $v\to0$ so $e^v\to1$; $w\to0$ so $\dfrac{e^w-1}w\to1$; and $\dfrac x{x+1}\to1$. The limit is $1$.

Remark. This is the idea behind the mean value theorem: $e^{u}-e^{v}\approx e^{v}(u-v)$ when $u$ and $v$ are close.