Why does \(\lim_{n \rightarrow \infty} nx(1-x^2)^n = 0\) for \(0<x<1\)?

[letmeknow]

Homework Statement

This should be easy but I can't find why.

Why is the following true for 0<x<1,

$$\lim_{n \rightarrow \infty}nx(1-x^2)^n = 0$$

Homework Equations

The Attempt at a Solution

I understand why (1-x^2)^n goes to zero, but the nx part is not bounded and seems to be going to infinity?

 

[lanedance]

can you use L'hopitals rule here?

 

[letmeknow]

I am not sure. I don't have much experience with L'Hospital's Rule so I will look it up.

 

[Mark44]

can you use l'hospital's rule here?

Not as the limit currently is written. L'Hopital's rule applies to quotients only, where both the numerator and denominator are approaching zero, or where both are approaching +/- infinity. To get this expression into a suitable form, instead of multiplying by n, you can divide by 1/n.

 

[lanedance]

another possibility for these types of limits is to write:

$$\lim_{n \rightarrow \infty} nx(1-x^2)^n = \lim_{n \rightarrow \infty} e^{\ln{nx(1-x^2)^n}}$$
and use the log laws to manipulate the expression, will have a go & see how I go

 

[lanedance]

hmm... the only way I manged this limit was to factor

$$\lim_{n \rightarrow \infty} nx(1-x^2)^n = \lim_{n \rightarrow \infty} (nx(1+x)^n)((1-x)^n)$$

then manipulating to make use of l'hospital's rule (for 0<x<1)
$$= \lim_{n \rightarrow \infty} \frac{(1-x)^n}{\frac{1}{nx(1+x)^n}} \rightarrow \frac{0}{0}$$

and differentiating numerator & denominator w.r.t. n gives the required limit, seems a little tricky though, not sure if I'm missing something...

 

[Mark44]

I don't think factoring the 1 - x^2 part makes life simpler. I would rewrite the original limit expression as $$x \lim_{n \rightarrow \infty} \frac{(1 - x^2)^n}{1/n}$$, and use L'Hopital's Rule on that. Both the numerator and denominator are approaching 0, so L'Hopital's Rule applies.