# Limit question

1. Mar 10, 2009

### letmeknow

1. The problem statement, all variables and given/known data
This should be easy but I can't find why.

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

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

2. Relevant equations

3. 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?

2. Mar 10, 2009

### lanedance

can you use L'hopitals rule here?

3. Mar 10, 2009

### letmeknow

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

4. Mar 10, 2009

### Staff: Mentor

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.

5. Mar 10, 2009

### 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

Last edited: Mar 10, 2009
6. Mar 10, 2009

### 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'hopitals 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...

7. Mar 11, 2009

### Staff: Mentor

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.