Let $A$ a matrix with real or complex entries. Proof that $\lim\limits_{n\rightarrow\infty}(E+\frac{A}{n})^n=e^A, E=$indentity. Let $A$ a matrix with real or complex entries. Prove that $\displaystyle\lim_{n\rightarrow\infty}\left(E+\frac{A}{n}\right)^n=e^A, E=$identity.
I thought of using the limit, but do not know where converges.
$$\displaystyle\lim_{n\rightarrow\infty}\sum_{i=0}^n \dfrac{n!}{(n-i)!n^ii!}$$
Hint: develop $\left(E + \dfrac{A}{n}\right)^n$ using Newton's Binomial and compare with $\displaystyle e^A = \sum \limits_{i=0}^\infty \dfrac{A^i}{i!}$
 A: I assume you are to show point-wise convergence. Let $|B|_{kj}$ denote the modulus of the $(k,j)$ entry of a matrix $B$. Let $2 \leq m \leq n$. Since $E$ commutes with $\frac{1}{n}A$ for any $n$ you can write (following the hint)
\begin{align}
  \left( E + \frac{A}{n} \right)^m
&= \sum_{i=0}^m \binom{m}{i}E^{m-i} \left( \frac{A}{n}\right)^i \\
&= \sum_{i=0}^m \binom{m}{i} \left( \frac{A}{n}\right)^i  \\
&= \sum_{i=0}^m \frac{A^i}{i!} \frac{m(m-1)\cdots (m-i+1)}{n^i}.
\end{align}
Since
$$
     0
\leq \frac{m\cdots (m-i+1)}{n^i}
\leq \frac{n\cdots (n-i+1)}{n^i}
\leq 1 
$$
when $i \in \{1,\cdots,m\}$,
$$
     \left| \sum_{i=0}^m \frac{A^i}{i!}\frac{n\cdots (n-i+1)}{n^i} \right|_{kj}
\leq \left| \left( E + \frac{A}{n} \right)^n \right|_{kj}
\leq \left| \sum_{i=0}^n \frac{A^i}{i!} \right|_{kj}
$$
(strictly speaking, the inequality on the left here can only be said to hold when $A$ has non-negative entries, which is too restrictive - I don't see any other way of using the hint though).
Take $n \to \infty$ and find
$$
     \left| \sum_{i=0}^m \frac{A^i}{i!} \right|_{kj}
\leq \liminf_{n \to \infty} \left| \left( E + \frac{A}{n} \right)^n \right|_{kj}
\leq \limsup_{n \to \infty} \left| \left( E + \frac{A}{n} \right)^n \right|_{kj}
\leq \left| e^A \right|_{kj}
$$
and then take $m \to \infty$. If you don't use the hint then the approaches suggested in the comments are good.
