Please evaluate the limit L given below. $$
L = \lim_{n \to \infty}\prod_{i = 1}^{n}\left(\, 1 + {i \over n^{2}}\,\right)
$$
Consider,
$$
L_{1} = \prod_{i = 1}^{n}\lim_{n \to \infty}\left(\,1 + {i \over n^{2}}\,\right)
$$
Q1. Is $L = L_{1}$ ?. Probably not but why can't I distribute the limit over product ?.
Q2. As I see it, each term in $L_{1}$ is individually $1$, therefore
$L_{1} = 1\cdot 1\cdots \infty$ times $ = 1$. But if I write
$L_{2} = \lim_{n \to \infty}\left(\, 1 + {1 \over n}\,\right)^{n}$ as
$$
\lim_{n \to \infty}\left(\,1 + {1 \over n}\,\right)\cdot
\lim_{n \to \infty}\left(\, 1 + {1 \over n}\,\right)\cdots \infty\ \mbox{times,}
$$
I get $1$ again because each term goes to $1$. However, it is equal to
$\,\mathrm{e}$. The mistake lies in the distribution of limit part. What is the mistake ?.
Please be elaborate because I can't get it ?.

$$\prod_{i=1}^{n}\left(1+\frac{i}{n^2}\right)=\exp\sum_{i=1}^{n}\log\left(1+\frac{i}{n^2}\right) $$
and the function $f(x)=\log(1+x)$ is bounded between $x$ and $x-\frac{x^2}{2}$ on the interval $[0,1]$, hence
$$\sum_{i=1}^{n}\log\left(1+\frac{i}{n^2}\right)=o(1)+\frac{1}{n^2}\sum_{i=1}^{n}i = \frac{1}{2}+o(1) $$
as $n\to +\infty$, and the wanted limit equals $\exp\left(\frac{1}{2}\right)=\color{red}{\large\sqrt{e}}$.
