Help solving the limit $\lim_{n\to \infty} \left(1+\frac{1}{n!}\right)^{2n}$ $$\lim_{n\to \infty} \left(1+\frac{1}{n!}\right)^{2n} $$
I've tried:
$$\lim_{n\to \infty} \left(1+\frac{1}{n!}\right)^{2n} = \lim_{n\to \infty} e^{\ln{\left(1+\frac{1}{n!}\right)^{2n}}} = \lim_{n\to \infty} e^{2n\ \ln{\left(1+\frac{1}{n!}\right)}}$$
But I don't know how to work with the factorial
 A: Since the exponential function is non-negative, we have for $x\geq 0$,
$$0\leq \int_0^x\int_0^y \exp(z)\,dz\,dy=\exp(x)-x-1. $$
Plugging into the inequality $1+x
\leq \exp(x)$ gives
$$1\leq \left(1+\frac{1}{n!}\right)^{2n}\leq \exp\left(2n/n!\right)=\exp\left(2/(n-1)!\right)\to 1$$
A: Hint: $$\ln(1+x)=x+O(x^2)$$ as $x\to 0$. (So this limit is equal to 1.)
or you can use inequality
$$\ln(1+x)\le x.$$
A: Here, I complete @milo's try: (we use $\ln(1+x)\sim_0x)
$$\lim_{n\to \infty} \left(1+\frac{1}{n!}\right)^{2n} = \lim_{n\to \infty} e^{\ln{\left(1+\frac{1}{n!}\right)^{2n}}} = \lim_{n\to \infty} e^{2n\ \ln{\left(1+\frac{1}{n!}\right)}}=\lim_{n\to \infty} e^{\frac{2n}{n!}}=e^0=1.$$
A: $$\lim_{n\to \infty} \left(1+\frac{1}{n!}\right)^{2n}=\lim_{n\to \infty}\left( \left(1+\frac{1}{n!}\right)^{n!}\right)^{\frac2{(n-1)!}}=e^0=1$$

Alternatively,
Let $$A=\left(1+\frac{1}{n!}\right)^{2n}$$
So, $$\log A=2n\log \left(1+\frac{1}{n!}\right)\text { using }\log(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots$$ 
which comes from the Taylor series which holds for $|x|<1$ as $n\to\infty, \frac1{n!}\to0\implies |\lim_{n\to\infty}\frac1{n!}|<1 $
$$=2n\left(\frac1{n!}-\frac{\frac1{(n!)^2}}2+\frac{\frac1{(n!)^3}}3+\cdots\right)$$
$$=\frac2{(n-1)!}-\frac1{n!(n-1)!}+\frac{2}{3(n-1)!(n!)^2}+\cdots$$
So, $$\lim_{n\to \infty}\log A=0\implies \lim_{n\to \infty}A=e^0=1 $$
