Limit of a Sequence of Functions

[jegues]

Homework Statement

Determine whether the sequence has a limit.

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

Homework Equations

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?

 

[╔(σ_σ)╝]

You could remove (1-x), sure.


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

 

[Mark44]

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

 

[jegues]

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

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

 

[jgens]

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

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

 

[jegues]

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

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,

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

Any suggestions/comments?

 

[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.

 

[jegues]

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.

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?

 

[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.

 

[jegues]

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.

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.

 

[jgens]

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.

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

 

[jegues]

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

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 that's why its indeterminant.

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

Could you show me some?

 

[jgens]

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

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.