need to show that lim at infinity of e^x is greater then x^p

**lllll** · February 10th 2009, 04:04 PM

I'm having a lot of trouble showing that that the $\lim_{x\rightarrow \infty} e^x > x^p \ \forall \ p \ \in \ \mathbb{R}$

I figure it has to be done by induction, so if I wanted to show that $e^x$ tends to infinity faster then $x^p$ I would have:

$\lim_{x\rightarrow \infty} e^x > x^1$ which is obviously true

so assume that

$\lim_{x\rightarrow \infty} e^x > x^p$ is true

then I would have $\lim_{x\rightarrow \infty} e^x > x^{p+1}= e^x>x^p\cdot x =\frac{e^x}{x}>x^p$ where the inequality holds only if $e^x$ tends to infinity at a faster rate then $x^p$

**ThePerfectHacker** · February 10th 2009, 05:20 PM

Originally Posted by lllll

I'm having a lot of trouble showing that that the $\lim_{x\rightarrow \infty} e^x > x^p \ \forall \ p \ \in \ \mathbb{R}$

I figure it has to be done by induction, so if I wanted to show that $e^x$ tends to infinity faster then $x^p$ I would have:

$\lim_{x\rightarrow \infty} e^x > x^1$ which is obviously true

so assume that

$\lim_{x\rightarrow \infty} e^x > x^p$ is true

then I would have $\lim_{x\rightarrow \infty} e^x > x^{p+1}= e^x>x^p\cdot x =\frac{e^x}{x}>x^p$ where the inequality holds only if $e^x$ tends to infinity at a faster rate then $x^p$

Hint: $e^x > 1 + x + \frac{x^2}{2!} + ... + \frac{x^k}{k!} + \frac{x^{k+1}}{(k+1)!}$ .

**NonCommAlg** · February 10th 2009, 05:25 PM

Originally Posted by lllll

I'm having a lot of trouble showing that that the $\lim_{x\rightarrow \infty} e^x > x^p \ \forall \ p \ \in \ \mathbb{R}$

I figure it has to be done by induction ??

p is a real number. so induction wouldn't work here! also why is this question in number theory section? your notation is not very good. you actually want to show that for sufficiently large values

of $x$ we have $e^x > x^p,$ which will follow if you show that $\lim_{x\to\infty}\frac{e^x}{x^p} = \infty.$ this is clear for $p \leq 0.$ for $p>0$ let $n=\lceil p \rceil$ and apply L'Hospital rule $n$ times to get the result.

**Jameson** · February 10th 2009, 05:30 PM

I moved it by mistake. I read "induction" and moved it without thinking. Luckily we have Moderators and other staff to watch after my actions.

**lllll** · February 10th 2009, 10:20 PM

Originally Posted by NonCommAlg

of $x$ we have $e^x > x^p,$ which will follow if you show that $\lim_{x\to\infty}\frac{e^x}{x^p} = \infty.$ this is clear for $p \leq 0.$ for $p>0$ let $n=\color{red}\lceil p \color{red}\rceil$ and apply L'Hospital rule $n$ times to get the result.

I'm not familiar with those brackets, do they have a particular property, or are they just like regular brackets?

**Jameson** · February 10th 2009, 11:31 PM

I think the brackets indicate the greatest integer function.

**ThePerfectHacker** · February 11th 2009, 02:12 PM

I will be explicit in my hint, I guess you did not see what I was trying to say. Let $x>1$ and $p\in \mathbb{R}, p>1$ . Pick $k\in \mathbb{Z}^+$ so that $k > p$ . If we can prove that $e^x > x^k$ for sufficiently large $x$ the proof will follow because $x^k > x^p$ . However, we know that $e^x > 1 + x + \tfrac{x^2}{2!}+...+\tfrac{x^{(k+1)}}{(k+1)!} > \tfrac{x^{k+1}}{(k+1)!}$ . However, if $x>(k+1)!$ we have $\tfrac{x^{(k+1)}}{(k+1)!} > x^k$ . Thus, we have shown that $e^x$ eventually overtakes $x^k$ and so it overtakes $x^p$ .

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