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$$\lim_{x\to\infty}{\frac{e^x}{x^a}}$$ For $a\in \Bbb R,$ find this limit.

I would say for $a\ge 0$ lim is equal to $\lim_{x\to\infty}{\frac{e^x}{a!x^0}=\infty}$ (from L'Hopital).

For $a<0$, lim eq. to $\frac{\infty}{0}$so lim doesnt exist. Is this correct?

Braindead
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lyme
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5 Answers5

3

Hint: Take the natural log of the expression. Then, you have

$$x - a \ln(x).$$

You can compute the limit without having to worry about whether $a$ is positive, negative, integer, or what not.

Braindead
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  • Why is it correct to take log ? – GorillaApe Feb 01 '15 at 12:59
  • If some expression $H(x)$ is positive, you can write $H(x) = exp(ln(H(x)))$. Using the fact that $exp$ and $ln$ are continuous, you can pass through the limit. Even if $H(x)->0+$, there is no problem. – Braindead Feb 01 '15 at 14:53
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Because $e^x > x$ for all $x$, $$\lim_{x \to \infty}\frac{e^x}{x}=\lim_{x \to \infty}\frac{1}{2}\left(\frac{e^{x/2}}{x/2}\right)e^{x/2} = \infty.$$ since $e^{x/2}/(x/2) > 1,$ and $e^{x/2} \to \infty$ as $x \to \infty$.

Then, it follows that $$\lim_{x \to \infty}\frac{e^x}{x^a}=\lim_{x \to \infty}\frac{1}{a^a}\cdot \left( \frac{e^{x/a}}{x/a}\right)^a = \infty$$

since we just showed that what is in parentheses approaches $\infty$ as $x \to \infty$, so the whole limit has to go to $\infty.$

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For $x>0$:

$\displaystyle \frac{e^x}{x^a} \geq \frac{1+x+x^2/2+x^3/3!+...+x^{\lfloor a\rfloor +2}/(\lfloor a\rfloor +2)!}{x^a} \geq \frac{x}{(\lfloor a\rfloor +2)!}$

When $x \rightarrow \infty$ ...

OBS: If $a$ is negative, it is easy to see that it goes to infinity.

Aloizio Macedo
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For any integer $n$, from the power series for $e^x$, $e^x > \frac{x^{n+1}}{(n+1)!}$ so $\frac{e^x}{x^n} > \frac{x}{(n+1)!} \to \infty$ as $x \to \infty$.

For any positive real $a$, choose $n > a$.

marty cohen
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Here is an elementary proof:

1) Claim 1: $\frac{x}{e^x}\to 0$. Since

$$\frac{1}{e}\frac{\lfloor x \rfloor}{e^{\lfloor x \rfloor}}\le \frac{x}{e^x}\le\frac{\lfloor x \rfloor+1}{e^{\lfloor x \rfloor+1}}e$$

and $n/q^n\to 0$ as $n\to \infty$ when $q>1$ (by ratio test). Thus $x/e^x\to 0$ as $x\to \infty$ by squeeze theorem.

2) Claim 2: $\frac{x^n}{e^x}\to 0$ for all $n\in \mathbb{Z}$. If $n\le 0$ the result is trivial. For $n> 0$, let $q = e^{1/n}>1$, so

$$\lim _{x\to \infty}\frac{x^{n}}{e^x}=\lim _{x\to \infty}\bigg(\frac{x}{q^x}\bigg)^n=\lim _{x\to \infty}\frac{x}{q^x}\cdot\lim _{x\to \infty}\frac{x}{q^x}\ldots \lim _{x\to \infty}\frac{x}{q^x}=0$$

3) Claim 3: $\frac{x^a}{e^x}\to 0$, $a\in \mathbb{R}$. For $x>1$ we have

$$0\le \frac{x^a}{e^x}\le \frac{x^{\lfloor a\rfloor+1}}{e^x}\to 0$$

Now given $N>0$ there is a $n_{0}$ such that $x^{a}/e^x<1/N$ for $x\ge n_0$. Thus $N<e^x/x^a$, i.e., $e^x/x^a \to \infty$ as $x\to \infty$.

Jose Antonio
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