Soit la suite
\[
x_n=e^{2\sqrt{n^2+1}-n}.
\]
Alors,
- \(\displaystyle\lim_{n\to\infty}x_n=1\)
- \(\displaystyle\lim_{n\to\infty}x_n=e\)
- \(\displaystyle \lim_{n\to\infty}x_n=+\infty\)
- \(\displaystyle\lim_{n\to\infty}x_n=0\)
\[\begin{aligned}
2\sqrt{n^2+1}-n&=
2n\sqrt{1+\frac{1}{n^2}}-n\\
&=
\underbrace{n}_{\to+\infty}\underbrace{\left(2\sqrt{1+\frac{1}{n^2}}-1\right)}_{\to
1\gt 0}
\end{aligned}\]