Limit of $\left(1+\frac{2}{n^2}\right)^n $ 
Compute$$\lim_{n\to\infty}\left(1+\frac{2}{n^2}\right)^n $$

I dont know how to do it without using continuity of exponential function
I mean I cannot do this:
$$\lim_{x\to\infty}a_n =a  \wedge  \lim_{x\to\infty}b_n =b\Rightarrow  \lim_{x\to\infty}{a_n}^{b_n} =a^b$$ 
 A: Hint: the expression equals $[(1+2/n^2)^{n^2}]^{1/n}.$
A: Use Bernoulli:
$$\left(1+\frac{2}{n^2}\right)^n= \frac{1}{\left(\frac{n^2}{n^2+2}\right)^n}= \frac{1}{\left(1-\frac{2}{n^2+2}\right)^n}$$
And by Bernoulli
$$\left(1-\frac{2}{n^2+2}\right)^n \geq 1-\frac{2n}{n^2+2}=\frac{n^2-2n+2}{n^2+2}$$
Therefore
$$1 \leq \left(1+\frac{2}{n^2}\right)^n \leq \frac{n^2+2}{n^2-2n+2}$$
A: You can consider the limit of the function (as opposed to sequence)
$$
f(x)=\left(1+\frac{2}{x^2}\right)^{\!x}
$$
or, better yet, the limit of its (natural) logarithm:
$$
\lim_{x\to\infty}\log\left(1+\frac{2}{x^2}\right)^{\!x}=
\lim_{x\to\infty}x\log\left(1+\frac{2}{x^2}\right)=
\lim_{t\to0^+}\frac{\log(1+2t^2)}{t}
$$
after the substitution $x=1/t$.
This is the derivative at $0$ of $g(t)=\log(1+2t^2)$ and
$$
g'(t)=\frac{4t}{1+2t^2}
$$
so $g'(0)=0$. Hence
$$
\lim_{x\to\infty}\log f(x)=0
$$
and therefore
$$
\lim_{x\to\infty}f(x)=e^0=1
$$
If the limit of the function exists, it is the same as the limit of the sequence.
A: Just another way.
Consider $$A=\left(1+\frac{2}{n^2}\right)^n\implies \log(A)=n\log\left(1+\frac{2}{n^2}\right)$$ Now, since $n$ is large, use Taylor series a round $x=0$ $$\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)\implies \log\left(1+\frac{2}{n^2}\right)=\frac{2}{n^2}-\frac{2}{n^4}+O\left(\frac{1}{n^6}\right)$$ $$\log(A)=n\log\left(1+\frac{2}{n^2}\right)=\frac{2}{n}-\frac{2}{n^3}+O\left(\frac{1}{n^5}\right)$$ Now, using $A=e^{\log(A)}$ an Taylor again $$A=1+\frac{2}{n}+\frac{2}{n^2}-\frac{2}{3 n^3}-\frac{10}{3
   n^4}+O\left(\frac{1}{n^5}\right)$$ which shows the limit and how it is approached.
All of the above was done for infinitely large values of $n$. However, it gives "good" results far even small values of $n$. For example, using $n=10$, the exact resulut would be $$A=\left(1+\frac{1}{50}\right)^{10}=\frac{119042423827613001}{97656250000000000}\approx 1.21899$$ while the approximation would lead to $\frac{1219}{1000}=1.21900$.
