# Limit of e^x

• October 12th 2013, 07:03 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, 07:50 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, 11:39 AM
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, 11:49 AM
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, 03:05 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, 03:47 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, 03:52 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, 06:08 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) :)
• May 5th 2015, 08:26 PM
limiTS
Re: Limit of e^x
I think you cannot use the Maclaurin series for $e^x$ which is about $x=0$ to analyze $e^x$ when $x\to\infty$. The expansion point (zero) is too far away from the limit point (infinity).

To solve the problem, you have to use L'Hospital's or just know that exponentials grow faster than polynomials.

This statement above is false: $\displaystyle\lim_{x\to\infty}\frac{x}{1-e^x}=-1$. The Maclaurin series approach gives an erroneous answer, because it cannot be used here.

Rather, the limit equals 0, as pointed out in other posts.
• May 5th 2015, 09:07 PM
Archie
Re: Limit of e^x
Quote:

Originally Posted by emakarov
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.

\begin{aligned} 0 &\lt {1 \over t} \lt t^{c-1} \qquad (t \gt 1,\, c \gt 0) \\ \int_1^x 0 \,\mathrm d t &\lt \int_1^x {1 \over t} \,\mathrm d t \lt \int_1^x t^{c-1} \,\mathrm d t \qquad (x \gt 1) \\ 0 &\lt \log x \lt {x^c - 1 \over c} \lt {x^c \over c} \\ 0 &\lt \log^a x \lt {x^{ac} \over c^a} \qquad (a \gt 0) \\ 0 &\lt {\log^a x \over x^b} \lt {x^{ac-b} \over c^a} \qquad (b \gt 0) \\ \end{aligned}
Since this is true for all $c \gt 0$ we may pick any $c = {b \over 2a}$ so that
$$0 \lt {\log^a x \over x^b} \lt {x^{-{b \over 2}} \over c^a} \\$$
And now, taking the limit as $x \to \infty$ gives
$$0 \le \lim_{x \to \infty} {\log^a x \over x^b} \le \lim_{x \to \infty} {x^{-{b \over 2}} \over c^a} = 0 \\ \implies \lim_{x \to \infty} {\log^a x \over x^b} = 0$$
Then, writing $x = \mathrm e^y$ we have
\begin{aligned} \lim_{x \to \infty} {\log^a x \over x^b} &= 0 \\ \lim_{y \to \infty} {\log^a \mathrm e^y \over \mathrm e^{by}} &= 0 \\ \lim_{y \to \infty} {y^a \over \mathrm e^{by}} &= 0 \\ \end{aligned}
\begin{aligned} \lim_{x \to \infty} {x^2 \over 1 - \mathrm e^x} &= \lim_{x \to \infty} {x^2 \mathrm e^{-x} \over \mathrm e^{-x} - 1} \\ &= \lim_{x \to \infty} {{x^2 \over \mathrm e^{x}} \over \mathrm e^{-x} - 1} \\ &= {0 \over 0 - 1} = 0 \end{aligned}
• May 5th 2015, 10:07 PM
votan
Re: Limit of e^x
Quote:

Originally Posted by Paze
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.

Is it x --> + infinity or - infinity. You could alternatively reason this way:

as x --> + infinity (1 - e^x) --> (- e^x)
At the limit the ratio varies as x^2/(-e^x) and the limit is 0 from the left

as x --> - infinity (1 - e^x) --> 1
At the limit the ratio varies as x^2 --> + infinity
• May 6th 2015, 04:57 AM
Archie
Re: Limit of e^x
Quote:

Originally Posted by limiTS
I think you cannot use the Maclaurin series for $e^x$ which is about $x=0$ to analyze $e^x$ when $x\to\infty$.

For a general function this is true, but the power series for $\mathrm e^x$ converges for all $x$, so in this case you can.