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'hopitals 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 independant 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