# Evaluating limit at infinity by Maclaurin series

• Sigma057
In summary, the problem is to evaluate the limit ## \lim_{x\to \infty } x^n e^{-x} ## using the Maclaurin expansion for ##e^{x}##. After dividing the numerator and denominator by ##x^n##, we are left with a series that can be simplified to ##\frac{1}{n!}\left(1+\frac{x}{n+1}+\frac{x^2}{(n+2) (n+1)}+\text{...}\right)##. By recognizing that n is a fixed finite constant, we can take the limit of the reciprocal sum and show that it goes to zero, thus solving the problem.
Sigma057

## Homework Statement

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}##.

## Homework 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.

## 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?

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 $x^n/e^x < 1/x$.

Sigma057 said:

## Homework Statement

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}##.

## Homework 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.

## 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?
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?

ehild said:
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?

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.

Sigma057 said:
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.
For ##x > 0##, all the terms in the series for ##e^x## are positive, so ##e^x > x^{n+k}/(n+k)!, ## hence
$$\frac{x^n}{e^x} < \frac{x^n}{x^{n+k}/(n+k)!} = (n+k)!/x^k.$$

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?

Sigma057 said:
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.
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?

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!

## 1. What is a Maclaurin series?

A Maclaurin series is a type of infinite series that represents a function as an infinite sum of terms that are powers of the variable x. It is named after the Scottish mathematician Colin Maclaurin.

## 2. How is a Maclaurin series used to evaluate limits at infinity?

A Maclaurin series can be used to evaluate limits at infinity by substituting x with 1/y and taking the limit as y approaches 0. This allows us to represent the function in terms of a polynomial, making it easier to evaluate the limit at infinity.

## 3. What is the formula for a Maclaurin series?

The formula for a Maclaurin series is given by f(x) = f(0) + f'(0)x + f''(0)x2/2! + f'''(0)x3/3! + ..., where f'(0), f''(0), f'''(0), etc. are the derivatives of the function f(x) evaluated at x = 0.

## 4. Can a Maclaurin series be used to evaluate any function at infinity?

No, a Maclaurin series can only be used to evaluate limits at infinity for functions that can be represented as an infinite sum of powers of x. This is known as a convergent Maclaurin series.

## 5. Are there any limitations to using a Maclaurin series to evaluate limits at infinity?

Yes, there are some limitations. A Maclaurin series can only be used to evaluate limits at infinity for functions that have a convergent Maclaurin series. Additionally, the series may only converge for certain values of x, so it is important to check the interval of convergence before using a Maclaurin series to evaluate a limit at infinity.

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