Proving the Limit of the Derivative Approaches Zero with a Function's Limit

[happyg1]

Hi,
I asked this question in a post a few days ago and got no response, so I thought I'd rephrase it and try again.
If I know that
$$\lim_{n\to\infty} f(n)=0$$
how can I prove that $$\lim_{n\to\infty} f'(n)=0$$?
My thoughts are: Since the function itsself is headed to zero that there will be a horizontal tangent line eventually, thus the derivative will be zero. I'm not sure if that's even reasonable.
Any help will be appreciated.
Thanks,
CC

 

[arildno]

Well, you can't, since the proposition is false!
Consider the function $f(n)=\frac{1}{n}\cos(e^{n})$ as a counter-example.

 

[happyg1]

now I'm confused even more if
$$\lim_{n\to\infty}f(n)=\frac{1}{n}cos(e^n)\neq 0$$
then
$$\lim_{n\to\infty}f'(n)=\lim(\frac{1}{n}(-sin(e^n)e^n)+cos(e^n)(-\frac{1}{n^2}))\neq 0$$
what am I missing?
according to the original thing I was trying to prove, this HAS to be true. I didn't want to clutter up my question with a lot of scary looking formulae. I think that's why no one responded.
CC

 

[happyg1]

ok ok ok (aching head...)
I don't think that's a counterexample, because the limit of the original function isn't zero, it oscillates and therefore has no limit.
PLEASE help me here
CC

 

[d_leet]

ok ok ok (aching head...)
I don't think that's a counterexample, because the limit of the original function isn't zero, it oscillates and therefore has no limit.
PLEASE help me here
CC

That limit is definitely zero, the top is always between -1 and 1 and the bottom goes to infinity so the limit is zero.

 

[arildno]

We have:
$$|f(n)|=\frac{|\cos(e^{n})|}{n}\leq\frac{1}{n}$$
Tends to zero..

 

[happyg1]

ok,
if you look at the original post, this is what I was trying to prove:
Given that b and c are not functions of n and $$\lim_{n\to\infty}\psi(n)=0$$
prove that

$$\lim_{n\to \infty}\left[ 1+\frac{b}{n} +\frac{\psi(n)}{n}\right]^{cn} = \lim_{n\to\infty} \left(1+\frac{b}{n}\right)^{cn}=e^{bc}$$
I see that the counter example works, but then this theorem from advanced calc is wrong. I just can't prove it. I'm missing something.
please take a look at "exponential function on the 2nd page of this forum. HELP
CC

 

[happyg1]

yes,
I read you there, but why then is this theorem true? Do I need to make some assumptions about this $$\psi(n)$$? And if so WHAT?
I have tried to use the definition of the derivative from the BEGINNING of the calc book, i.e.
$$\lim_{a\to\ 0}\frac{\psi(n+a)-\psi(n)}{a}$$
if I know that this is continuous then I can move the $$\lim_{a\to 0}$$ inside of the big limit up there, but I don't even know that.
What what what?
then IF that's true, don't i still have to apply L'hopital to that and go in another circle?
DANG
CC

 

[Cyrus]

I think I gave you wrong advice on the nonmonotonic limit. I am going to delete what I put, sorry.

 

[matt grime]

The full theorem appears true to me, though I don't see why you'd invoke l'hopital at all. It doesn't need differentiation. It is a straight forward epsilon type argument. Given any e>0 b/n+f(n)/n is bound between b/n-e and b/n+e for all n sufficiently large hence result follows. (Fill in the blanks.)

 

[arildno]

It is perhaps easiest to prove it by doing the following steps:
1. Assuming that $\lim_{n\to\infty}(1+\frac{b+\psi(n)}{n})^{n}$ exists, we'll have:
$$\lim_{n\to\infty}(1+\frac{b+\psi(n)}{n})^{cn}=(\lim_{n\to\infty}(1+\frac{b+\psi(n)}{n})^{n})^{c}$$

2. Since the exponential function is continuous, we have:
$$\lim_{n\to\infty}(1+\frac{b+\psi(n)}{n})^{n} =e^{\lim_{n\to\infty}n*log(1+\frac{b+\psi(n)}{n})}$$

3. Given that the right-hand side limits can be shown to exist, we have:
$$\lim_{n\to\infty}n*log(1+\frac{b+\psi(n)}{n})=\lim_{n\to\infty}n*log(1+\frac{b}{n})+\lim_{n\to\infty}n*log(1+\frac{\psi(n)}{n+b})$$

You may do the rest

 

[happyg1]

OK,
So the whole thing on the RHS there is still raised to the c power, right? and the$$e^{bc}$$ from the original equation is still on the FAR RHS..like this:
$$(\lim_{n\to\infty}n*log(1+\frac{b+\psi(n)}{n}))^c=(\lim _{n\to\infty}n*log(1+\frac{b}{n})+\lim_{n\to\infty }n*log(1+\frac{\psi(n)}{n+b}))^c = e^{bc}$$
I have a couple of questions. First, where did the$$(1+\frac{\psi(n)}{n+b})$$ come from? I can't get it to match up when I try to break it up. Second, if I take the limits of both of those terms, aren't they zero?So I will get something like $$0^c = e^{bc}$$ which isn't true.
This problem has really gotten me confused.
The reason I tried it with L'hopital is because that's what my professor suggested...
I know that this isn't as hard as I'm making it.
Any clarification will be GREATLY appreciated.
CC

 

[arildno]

$$((1+\frac{b}{n})+\frac{\psi}{n})=(1+\frac{b}{n})(1+\frac{\psi}{n(1+\frac{b}{n})})=(1+\frac{b}{n})(1+\frac{\psi}{n+b})$$

As for completing the argument, use that $$log(1+\epsilon)\approx\epsilon, \epsilon<<1$$

 

[happyg1]

OK,
Thanks for the algebra there. I was forgetting my n on the bottom after taking the 1+b/n out.
So what you're saying is that since $$log(1+\epsilon)\approx\epsilon$$ I should use that to get my terms to look a bit nicer. Let me see if I'm thinking like I need to be:
I will say that $$\left(\lim_{n\to \infty}(n*log(1+\frac{b}{n})+\lim_{n\to\infty}(1+\frac{\psi(n)}{b+n})\right)^c\approx(\epsilon n+\epsilon n)^c$$
then I raise it to the e power to get something like $$(e^{2n\epsilon})^c$$

am I even close?
I am still having trouble seeing how the two sides will match up.
My professor likes proofs to be super rigorous.
Thanks for your help.

 

[arildno]

nope, you have now, the approximate expression for the log sum:
$$n*\frac{b}{n}+n*\frac{\psi}{b+n}\approx{b}+\psi$$
That is, when exponentiating:

$$(e^{b+\psi})^{c}=e^{bc}e^{\psi{c}}$$

 

[happyg1]

Yes yes, I got that after taking a third look at what I did and putting the b/n and $$\frac{\psi}{n}$$ in for my $$\epsilon$$ terms. I see that the n's will cancel our in the first term. The second term is worrying me, though. We are assuming that the n*1/(b+n)is really close to 1 get that approximation, right?
Then when we exponentiate, are we still to take the limit to get that $$e^{\psi c}=1$$ ? We know that $$\lim_{n\to\infty}\psi(n)=0$$.
I guess my question is where in the proof do use that to get $$e^{\psi c}=1$$
Thankyou again.
CC

 

[arildno]

1. n/(b+n) is about 1 when n is big.
2. Since psi goes to zero, we have $$\lim_{n\to\infty}e^{\psi{c}}=e^{0}=1$$

 

[happyg1]

Thank you so much for all of your help on this. I actually understand it and the fog has been lifted.
Thank you thank you thank you.

 

[arildno]

You are welcome!

I guess it is the place to comment upon your original question, and state more precisely the conditions that must be met for the derivative to vanish at infinity if the function does.

Note that in the ugly oscillation function, the derivative is constantly CHANGING SIGNS.

If, however the function is (after some finite x-value) increasing (that is, its derivative non-negative) then the derivative is going to vanish at infinity.
Alternatively, of course, if the function is eventually decreasing, then the derivative will vanish as well.

 

[happyg1]

So if I somehow knew that the $$\psi(n)$$ function was monotone (up or down), then I could do the proof with L'hopital's rule. I wonder if I'm missing something in my givens or something I should have seen in the context of the question to lead me to assume that. This problem was given by my stats professor in addition to homework on limiting the moment generating function. Hmmmmmm...

 

[matt grime]

No, you cannot use l'hopital, where does it say anything about psi beign differentiable? Heck, all you needs is that psi is bounded and it works.