Calculating $\lim\limits_{n\to\infty} (2^n+1)^\frac{1}{n}$ using limit laws, standard limits, or otherwise

Question

I'm asked to find the $\lim\limits_{n\to\infty} (2^n+1)^\frac{1}{n}$. I'm really unsure as to how to do this, as there isn't really a way to expand without using binomial series.

Could anyone suggest some ideas?

Thanks :)

Answer 1

Hint: $$ (2^n)^{1/n} \le (2^n + 1)^{1/n} \le (2^{n+1})^{1/n} $$ and then apply the squeezing rule.

Answer 2

$$(1+2^n)^{1/n}=(2^n)^{1/n}\left(1+\dfrac1{2^n}\right)^{1/n}=2\left(\left(1+\dfrac1{2^n}\right)^{2^n}\right)^{1/n2^n}$$

Now $\lim_{n\to\infty}\left(1+\dfrac1{2^n}\right)^{2^n}=e$

and $\lim_{n\to\infty}\dfrac1{n2^n}=0$

Answer 3

$$\lim_{n\to\infty}\left(1+2^n\right)^{\frac{1}{n}}=\exp\left[\lim_{n\to\infty}\frac{\ln\left(1+2^n\right)}{n}\right]=$$

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Using l'Hôpital's rule:

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$$\exp\left[\ln(2)\cdot\lim_{n\to\infty}\frac{2^n}{1+2^n}\right]=\exp\left[\ln(2)\cdot\frac{1}{1+\lim_{n\to\infty}\frac{1}{2^n}}\right]=\exp\left[\ln(2)\cdot\frac{1}{1+0}\right]=2$$