Prove that "limit as n approaches infinity |x|^n/n! = 0 for all x"
We are finishing up Maclaurin series but I'm still having a hard time working problems out. I have a quiz on these types of problems in a couple days so anything will be helpful!
x^n = x*x*x*x... n times
n! = 1*2*3*... n times
No matter how big x gets, there will be infinitely many factors in the denominator greater than x.
Or, you may note that
∑x^n/n! = e^x
so the terms must converge to zero
To prove that the limit as n approaches infinity of |x|^n/n! is equal to 0 for all x, we can use the concept of the Maclaurin series expansion and the properties of limits.
First, let's start by expressing |x|^n/n! as a Maclaurin series expansion. The Maclaurin series of e^x is given by:
e^x = 1 + x + (x^2)/2! + (x^3)/3! + (x^4)/4! + ...
Now, notice that we can rewrite |x|^n as (x^n) or (x^(-n)) depending on whether x is positive or negative. Also, n! (factorial) can be expressed as n * (n-1) * (n-2) * ... * 2 * 1.
Therefore, we can rewrite |x|^n/n! as:
|x|^n/n! = (x^n)/n! for x >= 0
|x|^n/n! = ((-1)^n) * (|x|^n)/n! for x < 0
Now, to prove that the limit of |x|^n/n! is 0 as n approaches infinity, we need to show that the terms of the series get arbitrarily small as n increases.
For x >= 0:
As n increases, the numerator (x^n) increases or remains the same, while the denominator (n!) increases much faster because it includes multiplying terms from 1 to n. Therefore, the fraction (x^n)/n! becomes smaller and smaller as n approaches infinity, resulting in the limit approaching 0.
For x < 0:
Similarly, as n increases, both the numerator ((-1)^n) * (|x|^n) and the denominator (n!) increase, but the absolute value of the numerator (|x|^n) remains the same. Therefore, the fraction ((-1)^n) * (|x|^n)/n! becomes smaller and smaller as n approaches infinity, resulting in the limit approaching 0.
In both cases, since the terms of the series get arbitrarily small as n increases, the limit of |x|^n/n! approaches 0 for all real values of x.