1. Calculate

(i) $\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^{2}+1}}+\frac{1}{\sqrt{n^{2}+2}}+\ldots+\frac{1}{\sqrt{n^{2}+n}}\right)$;

(ii) $\lim _{x \rightarrow \infty}\left(1+\frac{1}{[x]}\right)^{x+1}$, where $[x]$ is the integer part of $x \in \mathbb{R}$.
1. 

(i) From the double inequality $\frac{n}{\sqrt{n^{2}+n}} \leq \frac{1}{\sqrt{n^{2}+1}}+\frac{1}{\sqrt{n^{2}+2}}+\ldots+\frac{1}{\sqrt{n^{2}+n}} \leq \frac{n}{\sqrt{n^{2}+1}}, n \in \mathbb{N}^{*} \quad$ 2p and applying the "squeeze" criterion, we deduce that the limit is 1.

(ii) $x \rightarrow \infty \Rightarrow [x] \rightarrow \infty$ we obtain the indeterminate form $1^{\infty}$ $1 p$

$\lim _{x \rightarrow \infty} \frac{x+1}{[x]}=1$

Finalization $\lim _{n \rightarrow \infty}\left(1+\frac{1}{[x]}\right)^{x+1}=e$.