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

Question

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: $$\lim_{x\to\infty}a_n =a \wedge \lim_{x\to\infty}b_n =b\Rightarrow \lim_{x\to\infty}{a_n}^{b_n} =a^b$$

Answer 1

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 $$1 \leq \left(1+\frac{2}{n^2}\right)^n \leq \frac{n^2+2}{n^2-2n+2}$$

Answer 2

Hint: the expression equals $[(1+2/n^2)^{n^2}]^{1/n}.$

Answer 3

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.

Answer 4

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 $\frac{1219}{1000}=1.21900$.