# Limit of e^x

• October 12th 2013, 08:03 AM
Paze
Limit of e^x
When working with a limit of e^x such as:

The limit as x approaches infinity of $\frac{x^2}{1-e^x}$

Can I simply state that it tends towards 1/((e^inf)/inf) and therefore to 0, or is the limit not yet defined since it tends towards inf/inf below 1? Thanks.
• October 12th 2013, 08:50 AM
emakarov
Re: Limit of e^x
Why do you think that 1/((e^inf)/inf) is 0? In general, you can't compare infinities.
• October 12th 2013, 12:39 PM
Paze
Re: Limit of e^x
Quote:

Originally Posted by emakarov
Why do you think that 1/((e^inf)/inf) is 0? In general, you can't compare infinities.

So I'll need l'hopital right?
• October 12th 2013, 12:49 PM
emakarov
Re: Limit of e^x
Yes, you could use L'Hopital's rule. To find limits, you need some database of standard facts, e.g., $\frac{a_nx^n+\dots+a_0}{b_nx^n+\dots+b_0}\to\frac{ a_n}{b_n}$ as $x\to\infty$ provided $a_n\ne0$ and $b_n\ne0$. The fact that the $\lim_{x\to\infty}\frac{x^n}{a^x}=0$ where $a>1$ is one of those standard facts, but one has to prove it for the first time.
• October 12th 2013, 04:05 PM
HallsofIvy
Re: Limit of e^x
You could also use the MacLaurin series for $e^x$, $1+ x+ \frac{x^2}{2}+ \frac{x^3}{6}+ \cdot\cdot\cdot$ so that $1- e^x= -(x+ \frac{x^2}{2}+ \frac{x^3}{6}+ \cdot\cdot\cdot}$.

Now you have $\frac{x^2}{1- e^x}= -\frac{x^2}{x+ \frac{x^2}{2}+ \frac{x^3}{6}+ \cdot\cdot\cdot}$.

Divide both numerator and denominator by $x^2$ to get $-\frac{1}{\frac{1}{x}+ \frac{1}{2}+ \frac{x}{6}+ \cdot\cdot\cdot}$ which goes to infinity as x goes to infinity. Notice that, by this argument, $\lim_{x\to\infty} \frac{x}{1- e^x}= 1$.
• October 12th 2013, 04:47 PM
Paze
Re: Limit of e^x
Quote:

Originally Posted by HallsofIvy
You could also use the MacLaurin series for $e^x$, $1+ x+ \frac{x^2}{2}+ \frac{x^3}{6}+ \cdot\cdot\cdot$ so that $1- e^x= -(x+ \frac{x^2}{2}+ \frac{x^3}{6}+ \cdot\cdot\cdot}$.

Now you have $\frac{x^2}{1- e^x}= -\frac{x^2}{x+ \frac{x^2}{2}+ \frac{x^3}{6}+ \cdot\cdot\cdot}$.

Divide both numerator and denominator by $x^2$ to get $-\frac{1}{\frac{1}{x}+ \frac{1}{2}+ \frac{x}{6}+ \cdot\cdot\cdot}$ which goes to infinity as x goes to infinity. Notice that, by this argument, $\lim_{x\to\infty} \frac{x}{1- e^x}= 1$.

According to calculators and l'hopitals, we have it tending towards 0. Which one is correct...?
• October 12th 2013, 04:52 PM
Plato
Re: Limit of e^x
Quote:

Originally Posted by Paze
According to calculators and l'hopitals, we have it tending towards 0. Which one is correct...?

In both cases the limit is $0$.
I myself find Prof. Ivey's post confusing. But his point is valid: the series approach gives that limit.
• October 12th 2013, 07:08 PM
Paze
Re: Limit of e^x
Quote:

Originally Posted by Plato
In both cases the limit is $0$.
I myself find Prof. Ivey's post confusing. But his point is valid: the series approach gives that limit.

Interesting. I'm going to expand on this idea for my assignment (I hate regurgitating trivial information from a blackboard and I don't care if I get downgraded for veering off course) :)