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 ?.

In the formula $$L_1=\prod_{i=1}^{\style{color:#f00}{n}}\lim_{n\to \infty}(a+\frac i {n^2})$$ the $n$ marked with red does not denote anything, it is neither bound to an expression nor is it predefined, whereas in the definition of $L$ is is bounded by the $\lim_{n\to \infty}$. It is the same as if you would take $$\prod_{i=0}^1\prod_{j=1}^ij$$ and write $$\prod_{j=1}^{\style{color:red}{i}}\prod_{i=0}^1j$$ There is simply no rule that enables you to even state that they are equal, not alone proving that they are equal.
