Calculating $\lim_{x\to\infty} \frac{x^{5}}{2^{\sqrt{x}}}$

Question

I'm interested in calculating

$$\lim_{x\to\infty}\frac{x^{5}}{2^{\sqrt{x}}}$$

I'm thinking of using L'Hopital's Rule here, which gives:

$$\lim_{x\to\infty}\frac{5x^{4}\sqrt{x}}{\ln(2)\cdot 2^{\sqrt{x}-1}}$$

But it doesn't look so great, is this the wrong approach?

Answer 1

This is essentially boils down to making the right substitutions. First we try $x \rightarrow x^2$ to get rid of the square root. Now we have $\frac{x^{10}}{2^x}.$ At this point, you can finish the proof by using the fact that exponentials always outpace polynomials.

Your job is rigorize my 2 statements to an acceptable level. If this is enough for you, then that's fine. If not, I suggest providing a proof for the last step. Either way, good luck!

Note: Now that I think about it, you can prove that $\forall a, n > 1$, $\frac{x^n}{a^x}$ tends to $0$ by strong induction and L'Hopitals rule.

Answer 2

Display name beat me to the punch, but I wanted to add, your approach will work. Don't forget that you can combine $x^4$ and $\sqrt{x}$ to make $x^{4.5}$. Each time you perform L'Hopital's rule, you'll decrease the power by $0.5$, until you obtain an expression that is a nasty multiple of $2^{-\sqrt{x}}$, which clearly approaches $0$.

Answer 3

Note that eventually for $x\to+\infty$

$${2^{\sqrt{x}}}\ge x^{6}$$

then eventually we have that

$$\frac{x^5}{2^{\sqrt{x}}}\le\frac{x^5}{x^6}=\frac1x \to 0 $$

Indeed It is a standard result that $\forall c$ real

$$a_n=\frac{n^c}{2^n}\to 0$$

indeed by ratio test

$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^c}{2^{n+1}}\frac{2^n}{n^c}=\frac12\left(\frac{n+1}{n}\right)^c\to \frac12<1$$

and from here by squeeze theorem can be shown that also for $y\to+\infty$ real

$$\frac{y^c}{2^y}\to 0$$

and setting $y=\sqrt x\to +\infty$ and $c=10$ we obtain the given limit.

Answer 4

Let $y: =√x$ and consider $y \rightarrow \infty.$

$0 \le F(y):= \dfrac {y^{10}}{2^y}.$

$2^y =e ^{y\log2}.$

$F(y)= \dfrac{y^{10}}{e^{y\log2}}.$

Note: $0 \lt \log 2 \lt 1. $

Set $a:=\log 2.$

$e^{ay} = 1 + (ay) + ...\dfrac{(ay)^{11}}{11!} +...\ge$

$\dfrac{(ay)^{11}}{11!}.$

$0 \le F(y) \le \dfrac{(11!)y^{10}}{a^{11}y^{11}}=\dfrac{(11!)}{a^{11}}\dfrac{1}{y}.$

Squeeze.

The $ \lim_{y \rightarrow \infty} F(y)$ is?

Answer 5

This all boils down to the proof that exponential functions tend to infinity much faster than polynomials as the argument tends to infinity or equivalently the polynomials tend to infinity much faster than logarithmic functions.

In symbols we have for any $k>0, a>1$ $$\lim_{x\to\infty}\frac{x^{k}}{a^x}=0=\lim_{x\to\infty}\frac{\log x} {x^k}\tag{1}$$ The above results are pretty standard and can be used anywhere just the way we use the more famous limit formula $$\lim_{x\to 0}\frac{\sin x} {x} =1\tag{2}$$ Your question is now easily solved by using the substitution $x=t^2$ to get $$\lim_{x\to \infty} \frac{x^5}{2^{\sqrt{x}}}=\lim_{t\to\infty} \frac{t^{10}}{2^t}=0$$ In case you want a proof of the formula $(1)$ it is an easy consequence of the properties of exponential and logarithmic functions.

We can use the fundamental inequality $\log x\leq x-1$ to get $$\log x=\frac{1}{p}\log x^p\leq \frac{x^p-1}{p}<\frac{x^p}{p}$$ where $p$ is a number such that $0<p<k$ and then for $x>1$ we have $$0<\frac{\log x} {x^k} <\frac{1}{px^{k-p}}$$ and by Squeeze Theorem we get $$\lim_{x\to\infty} \frac{\log x} {x^k} =0$$ For the exponential limit just use the substitution $a^x=t$ so that $x=(\log t) /\log a$ and $$\frac{x^k} {a^x} =(\log a) ^{-k} \cdot\left(\frac{\log t} {t^{1/k}}\right)^{k}\to (\log a) ^{-k} \cdot 0=0$$