# Evaluating limit at infinity by Maclaurin series

1. Apr 10, 2017

### Sigma057

1. The problem statement, all variables and given/known data
I've begun going through Boas' Math Methods in the Physical Sciences and am stuck on problem 1.15.25. The problem is to evaluate
$\lim_{x\to \infty } x^n e^{-x}$
By using the Maclaurin expansion for $e^{x}$.

2. Relevant equations
We know the Maclaurin expansion for the exponential function to be
$1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\text{...}$
We are also given the hint to divide the numerator and denominator by $x^n$ before taking the limit.

3. The attempt at a solution
Taking the hint I proceed
$e^{-x} x^n=\frac{x^n}{e^x}=\frac{1}{\frac{e^x}{x^n}}=\frac{1}{\frac{1}{x^n}\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\text{...}\right)}=\frac{1}{\frac{1}{x^n}+\frac{x}{x^n}+\frac{x^2}{2 x^n}+\frac{x^3}{6 x^n}+\frac{x^4}{24 x^n}+\text{...}}=\frac{1}{\frac{1}{x^n}+\frac{1}{x^{n-1}}+\frac{1}{2 x^{n-2}}+\frac{1}{6 x^{n-3}}+\frac{1}{24 x^{n-4}}+\text{...}}$
I really don't see how this helps me as I let x go to infinity.
Any suggestions?

2. Apr 10, 2017

### stevendaryl

Staff Emeritus
I had Ralph Boas (husband of the author of that book) for a math course once. He was a very sweet old guy.

I'm not exactly sure how she wants you to solve the problem, but if you could prove that, for $x$ large enough, $e^x > x^{n+1}$, then it would follow that $x^n/e^x < 1/x$.

3. Apr 10, 2017

### ehild

The denominator is $\frac{1}{x^n}+\frac{1}{x^{n-1}}+\frac{1}{2 x^{n-2}}+\text{...}+\frac{1}{n!}+\frac{1}{(n+1)!}x+\frac{1}{(n+2)!}x^2+\text{...}$
What happens if x goes to infinity?

4. Apr 12, 2017

### Sigma057

As x goes to infinity, $\frac{1}{x^n}, \frac{1}{x^{n-1}}, \frac{1}{2 x^{n-2}},...$ go to zero leaving us to evaluate the rest of the series $\frac{1}{n!}+\frac{1}{(n+1)!}x+\frac{1}{(n+2)!}x^2+\text{...}$

The only step I can see from here is to factor out an n!.
$\frac{1}{n!}+\frac{1}{(n+1)!}x+\frac{1}{(n+2)!}x^2+\text{...} = \frac{1}{n!}+\frac{x}{(n+1) n!}+\frac{x^2}{(n+2) (n+1) n!}+\text{...} = \frac{1}{n!}\left(1+\frac{x}{n+1}+\frac{x^2}{(n+2) (n+1)}+\text{...}\right)$
And once again I'm stuck as I know of no way to evaluate this.

5. Apr 12, 2017

### Ray Vickson

For $x > 0$, all the terms in the series for $e^x$ are positive, so $e^x > x^{n+k}/(n+k)!,$ hence
$$\frac{x^n}{e^x} < \frac{x^n}{x^{n+k}/(n+k)!} = (n+k)!/x^k.$$

6. Apr 12, 2017

### vela

Staff Emeritus
Seems like you're making this more complicated than it needs to be. Doesn't
$$\lim_{x\to\infty} \frac{1}{n!} + \frac{1}{(n+1)!} x + \cdots$$ diverge?

7. Apr 12, 2017

### ehild

Remember that n is a finite number.
$1+\frac{x}{n+1}+\frac{x^2}{(n+2) (n+1)}+\text{...}>1+\frac{x}{n+1}$ which goes to infinity if x-->∞. What is the limit of its reciprocal?

8. Apr 12, 2017

### Sigma057

Thank you all! The fact that n is a fixed finite constant was the fundamental concept I wasn't using in my solution. I now can see that once I take the limit of the reciprocal sum it goes to zero =)

You have no idea how frustrating this problem has been (for no good reason). Thank you al for sharing your insights with a fresh pair of eyes!