Can't understand the limit solution to $\lim_{n\to\infty}\sqrt[n]{2^n-n^2}$

Question

I trying to solve the question:

$$\lim_{n\to\infty}\sqrt[n]{2^n-n^2}$$

I know that i can proof $2^n>n^2$ in induction
and i calculated the limit as "$∞^0$" that equals to $1$
but then i saw that the answer should be 2.

How is it 2?

Answer 1

$$(2^n-n^2)^{1/n}=(2^n)^{1/n}\left\{\left(1-\dfrac{n^2}{2^n}\right)^{-2^n/n^2}\right\}^{-n/2^n}$$

Now $\lim_{n\to\infty}\left(1-\dfrac{n^2}{2^n}\right)^{-2^n/n^2}=e$

$\lim_{n\to\infty}\dfrac n{2^n}=?$

Answer 2

Hint:

$$ \sqrt[n]{2^n-n^2}=2\cdot \sqrt[n]{1-\frac{n^2}{2^n}}$$

Answer 3

You need to use the result that $n^{2}/2^{n - 1} \to 0$ as $n \to \infty$ and therefore for large values of $n$ we have $$2^{n - 1} < 2^{n} - n^{2} < 2^{n}$$ and hence $$\frac{2}{2^{1/n}} < \sqrt[n]{2^{n} - n^{2}} < 2$$ and applying Squeeze theorem we get the desired limit as $2$.

Answer 4

It amounts to finding the limit of the log: $\;\dfrac1n\ln(2^n-n^2)$.

The easiest way to find the limit uses equivalents:

$2^n-n^2\sim_\infty 2^n$. As $\;2^n-n^2$ doesn't approach $1$ for $n\to\infty$, we can take the log, so that $\; \dfrac1n\ln(2^n-n^2)\sim_\infty \ln 2^n=n\ln2 $, hence $$\ln\Bigl(\sqrt[n]{2^n-n^2}\Bigr)\sim_\infty\frac1n\cdot n\ln2=\ln2.$$ Thus $\;\sqrt[n]{2^n-n^2}\xrightarrow[n\to\infty]{}2$.

Answer 5

First you can collect the "dominant" term under the $n$ root:

$\lim_{n\rightarrow +\infty}\sqrt[n]{2^n-n^2}=\lim_{n\rightarrow +\infty}\sqrt[n]{2^n\left(1-\dfrac{n^2}{2^n}\right)}$

Obviously, you can also rewrite as:

$2\cdot \lim_{n\rightarrow +\infty}\sqrt[n]{\left(1-\dfrac{n^2}{2^n}\right)}$

Now; for all $n\geq 1$, the term $\left(1-\dfrac{n^2}{2^n}\right)$ is positive, so you can use one of the fundamental properties of the logarithms: if $x>0$ then $x=e^{\ln(x)}$; obtaining:

$2\cdot \lim_{n\rightarrow +\infty}\sqrt[n]{\left(1-\dfrac{n^2}{2^n}\right)}= 2\cdot \lim_{n\rightarrow +\infty}{e^{\ln\left(\sqrt[n]{1-\dfrac{n^2}{2^n}} \right)}}$

Using basic properties of the logarithm, rewrite as:

$2\cdot \lim_{n\rightarrow +\infty}{e^{\frac{1}{n} \ln\left(1-\frac{n^2}{2^n}\right)}}$

Now, when $n\rightarrow +\infty$, $\frac{n^2}{2^n}\rightarrow 0$, so you can use asymptotic expansion of the logarithm:

$\ln\left(1-\frac{n^2}{2^n}\right)\sim -\frac{n^2}{2^n}$

Your limit becomes:

$2\cdot \lim_{n\rightarrow +\infty}{e^{-\frac{1}{n}\frac{n^2}{2^n}}}= 2\cdot \lim_{n\rightarrow +\infty}{e^{-\frac{n}{2^n}}}$

From elementary limits, you know that $-\frac{n}{2^n}\rightarrow 0$ when $n\rightarrow +\infty$; so $\lim_{n\rightarrow +\infty}{e^{-\frac{n}{2^n}}}=e^0=1$

From this, you have that your limit is $2$.

Hope this help.