Does $\lim_{n \to \infty} a_n = \alpha$ imply $\lim \left(1+ \frac {a_n}{n} \right) =e^\alpha$ for real value Does $\lim_{n \to \infty}  a_n = \alpha$   imply $\lim_{n \to \infty} \left(1+ \frac {a_n}{n} \right)^n =e^\alpha$?
I'm only considering real value
and also, does the value of $\alpha$ makes a difference?\
I tried considering $$\left|\left(1+ \frac {a_n}{n} \right)^n-e^\alpha\right|\leq\left|\left(1+ \frac {a_n}{n} \right)^n-\left(1+ \frac {\alpha}{n} \right)^n\right|+\left|\left(1+ \frac {\alpha}{n} \right)^n-e^\alpha\right|$$
I know that for any fixed $\alpha$, $\left|\left(1+ \frac {\alpha}{n} \right)^n-e^\alpha\right|$ tends to zero.
But I don't know how to deal with $$\left|\left(1+ \frac {a_n}{n} \right)^n-\left(1+ \frac {\alpha}{n} \right)^n\right|$$, I tried binomial expansion but didn't get anywhere. Is this where the value of $\alpha$ come to matter？
Thanks
 A: HINT: If $a_n\to \alpha > 0$ as $n\to\infty$, then $b_n := na_n\to\infty$ as $n\to\infty$. Therefore, make the substitution $n \mapsto na_n$ in the limit, which yields the following equality:
$$\lim_{n\to\infty} \bigg(1+\frac{a_n}{n}\bigg)^n = \lim_{n\to\infty} \bigg(1+\frac{a_n}{na_n}\bigg)^{na_n} = \lim_{n\to\infty} \bigg(1+\frac{1}{n}\bigg)^{na_n}$$
Does this help?
A: It requires breaking up the expansion of $e^a$ into two parts depending on $\epsilon$.  Let $N_1$ be defined by $\sum\limits_{k=N_1+1}^\infty \frac{a^k}{k!}\lt \epsilon$.  Then the binomial expansion can be used on the sum to $N_1$.  Let $N_2$ be defined by $\sum\limits_{k=0}^{N_1} \binom{n}{k}|a_n^k-a^k|\lt \epsilon$ for $n\gt N_2$.  Let $N=max(N_1,N_2)$  Expansion differences should be $\lt 2\epsilon$ for $n\gt N$.
To handle the tails for $a_n$ would require a little work showing $|\sum\limits_{k=N_1}^n\binom{n}{k}a_n^k-\sum\limits_{k=N_1+1}^\infty \frac{a^k}{k!}|\lt \sum\limits_{k=N_1+1}^\infty \frac{a^k}{k!}$ for $n\gt N$.
