How to calculate limit of nth root of $(1+2^n)$

Question

Somehow I need to get $2$ as the answer but when I write

$$\displaystyle\lim_{n\to\infty}(1+2^n)^{\displaystyle \lim_{n\to\infty}{\frac{1}{n}}}=\infty^0=1$$

You can't take an expression like $\lim_{n\to \infty}(1+2^n)^{1/n}$ and take the limit of each part separately. That only works if both the exponent and the base converge separately to something finite, which is not the case here. — Arthur, Jan 29 '18 at 10:25
You mean I should write lim(1)^(1/n) + lim(2^n)^(1/n) and then I get 0+2 =2 — Jo King, Jan 29 '18 at 10:26
$nth$ root of $(1+2^n)$ means $(1+2^n)^(1/n)$ you can’t distribute power to each term since you are adding. — Harry, Jan 29 '18 at 10:30

score 5 · Accepted Answer · answered Jan 29 '18 at 10:30

5

We have $$\lim_{n\to\infty}(1+2^n)^{1/n}=2\lim_{n\to\infty}\left(1+\frac1{2^n}\right)^{1/n}=2\cdot1^0=2$$

answered Jan 29 '18 at 10:30

TheSimpliFire

score 2 · Answer 2 · answered Jan 29 '18 at 14:47

$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n}=L \implies\lim_{n\to\infty} \sqrt[n]{a_n}=L$$

let $a_n=1+2^n$ since

$$\frac{a_{n+1}}{a_n}=\frac{1+2^{n+1}}{1+2^n}=\frac{1+2\cdot 2^{n}}{1+2^n}=\frac{\frac1{2^n}+2}{\frac1{2^n}+1}\to2 $$

we have

$$\lim_{n\to\infty} \sqrt[n]{a_n}=\lim_{n\to\infty}(1+2^n)^{\frac1n}=2$$

2 Answers2