Proving Limit of Exponential Function is Zero

[happyg1]

Hi,
I'm working on this:
Given that $$\lim_{n \to \infty} \psi(n)=0$$ and that b and c do not depend upon n, prove that:
$$\lim_{n\to \infty}\left[ 1+\frac{b}{n} +\frac{\psi(n)}{n}\right]^{cn} = \lim_{n\to\infty} \left(1+\frac{b}{n}\right)^{cn}=e^{bc}$$
So far, I've taken the natural log of both sides, moved the cn into the bottom and applied l'hospital's rule. I get:
$$\lim_{n\to\infty}\frac{\frac{1}{1+\frac{b}{n} +\frac{\psi(n)}{n}}\left(\frac{-b}{n^2} + \frac{\psi'(n)}{n}-\frac{\psi(n)}{n^2}\right)}} {\frac{-1}{c n^2}}}$$$$=\lim_{n\to \infty}bc$$
which breaks down to:
$$\lim_{n\to\infty}\frac{1}{1+\frac{b}{n}+\frac{\psi(n)}{n}}\left(-cn\psi'(n)-c\psi(n)+bc)\right)$$
If the limit of a function goes to zero, how do we prove that it's derivative goes to zero?
I'm not sure where to go now, because I don't know what to do with $$\psi'(n)$$ how can I prove that it's zero? If it IS zero, then the whole thing falls out nicely.
Thanks,
CC

 

[quasar987]

There is some problem with your LaTeX.

Is psi a wave function by any chance?

 

[happyg1]

yea, I tried to fix the latex a little bit. For some reason, it's not centering my fraction on the bottom.That whole mess is over $$\frac{-1}{c^2 n^2}$$. No, the $$\psi(n)$$ isn't a wave function. This is for a statistics class. I think the only thing I'm supposed to worry about is that it goes to 0 as n goes to infinity. I also forrgot my limit in front of my giant fraction. I will try and fix it.
CC

 

[happyg1]

There. It's a little better. I hope you can tell what I mean. I can't figure out why the bottom fraction is way over on the left like that. That WHOLE thing on the left equals $$\lim_{n\to\infty}bc$$...which is just bc...anyway, any pointers will be appreciated. I am stuck stuck stuck.
CC

 

[Deau]

Well I know this post is 3 years old so you likely no longer need the answer but who knows, maybe someone will make some use of it.

I will try and type it with LaTeX code but I am not positive it will show as intended as this is my first use of these forums.

The basic principle used here will be the well known exponential convergence which is

$$\lim_{n \to \infty} (1 + \frac{x}{n})^{n} = e^{x}$$

Now looking at your problem. Since c is independent from n, we have

$$\lim_{n\to \infty}\left[ 1+\frac{b}{n} +\frac{\psi(n)}{n}\right]^{cn} & = & {\left[ \lim_{n\to \infty}\left[ 1+\frac{b}{n} +\frac{\psi(n)}{n}\right]^{n} \right]}^{c}$$
$$= {\left[ \lim_{n\to \infty}\left[ 1+\frac{b+\psi(n)}{n}\right]^{n} \right]}^{c}$$
$$= {\left[ \lim_{n\to \infty} e^{b+\psi(n)} \right]}^{c}$$
$$= {\left[ e^{b} \times \lim_{n\to \infty} e^{\psi(n)} \right]}^{c}$$
$$= {\left[ e^{b}\times e^{0} \right]}^{c}$$
$$= e^{bc}$$



Vincent
Graduate math student