Limit of a Sequence of Functions

1. Sep 15, 2010

jegues

1. The problem statement, all variables and given/known data

Determine whether the sequence has a limit.

$$f_{n}(x) = nx^{n}(1-x)$$ for $$0 \leq x \leq 2$$

2. Relevant equations

3. The attempt at a solution

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

I can take the (1-x) out and I'll have,

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

Therefore, there is no limit function.

Is this correct?

Last edited: Sep 15, 2010
2. Sep 15, 2010

╔(σ_σ)╝

You could remove (1-x), sure.

However, the limit as n goes to infinity may not diverge when $$0 \leq x \leq 1$$

3. Sep 15, 2010

Staff: Mentor

For the domain for your sequence of functions, don't you mean
$$0 \leq x \leq 2$$
?

4. Sep 15, 2010

jegues

Yes, I'll edit it now. Still not sure if my method for concluding that there is no limit is valid or not.

5. Sep 16, 2010

jgens

It's not. You need to consider a couple of cases.

6. Sep 16, 2010

jegues

Well if x = 1,

$$f_{n}(1) = n1^{n}(1-1)$$

then

$$lim_{n \rightarrow \infty} f_{n}(1) = 0$$

Well what about when x != 1,

$$f_{n}(x) = nx^{n}(1-x)$$

So,

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

Therefore there is no limiting function.

Is this enough? I'm not sure how to properly deal with,

$$lim_{n \rightarrow \infty} nx^{n}$$

Last edited: Sep 16, 2010
7. Sep 16, 2010

jgens

Why do you think that $\lim_{n \to \infty}nx^n = \infty$? It's not generally true, so it's not quite as simple as you seem to suggest.

8. Sep 16, 2010

jegues

I can't think of how to deal with this limit, this is the only thing I can come up with.

Can you make me some suggestions and I'll try them out?

9. Sep 16, 2010

jgens

Sure. You've already considered the case when $x=1$, now consider when $0 \leq x < 1$ and $1 < x \leq 2$. Treat the cases separately.

10. Sep 16, 2010

jegues

Okay for the first case,

$0 \leq x < 1$

$$lim_{n\rightarrow \infty} nx^{n}$$

In this particular case x^n term will go to 0 and the n term will go to infinity, so the limit does not exist.

$1 < x \leq 2$

In this case, the n term will go to infinity and the x^n term will go to infinity, so the limit does not exist either.

Is that all have I have to state? I don't know how else I can show this.

11. Sep 16, 2010

jgens

No. How does this justify that the limit doesn't exist?

12. Sep 16, 2010

jegues

I've already told you, I don't know what else to say about it.

I don't know how I can change nx^n into something that is clearly a function or clearly indeterminant.

The closest I can get is to tell you that it is 0 times infinity and thats why its indeterminant.

I know that isn't correct, but I can't see any other options.

Could you show me some?

13. Sep 16, 2010

jgens

You never said anything about the particular case that you were working on (0 < x < 1) other than that the limit doesn't exist. In my opinion, it would be better to ask for help than say something incorrectly. If you don't know how to handle this limit when 0 < x < 1, then ask. You never did that.

Anyway, with that aside, there are a couple of tricks that you can use to evaluate that limit for 0 < x < 1. I might do it by noting that

$$nx^n = e^{\log{(nx^n)}} = e^{\log{(n)+n\log{(x)}}$$

$$\lim_{x \to \infty}e^{f(x)} = e^{\lim_{x \to \infty}f(x)}$$

Granted, the last equality only holds under certain circumstances, but you should be able to figure out whether or not it's valid.