1. ## proof on limits

Problem
Let f:R-->R be a differentiable function. Prove that if lim x-->infinity f(x) and lim x-->infinity f'(x) both exist and are finite, then lim x--> infinity f'(x) = 0.

2. You should notice that for some $\displaystyle \epsilon>0$, $\displaystyle |f(x)-f(t)|<\epsilon$ whenever $\displaystyle x,t>X$ for some $\displaystyle X$. Now look at the definition of derivative: $\displaystyle f'(x) = \lim_{t->x}{\frac{f(t)-f(x)}{t-x)}}$. This number exists by differentiability of f.
Now, for another given $\displaystyle \epsilon>0$, you need to prove that $\displaystyle |f'(x)|<\epsilon$ using above relations.

Hope this helps

3. i don't see what you're getting at?

i don't see how you end up at lim x-->infinity f'(x) =0

4. Forget what I said.
I have another idea in mind.
Use MVT on the interval $\displaystyle [x,x+1]$:
we have: $\displaystyle f(x+1)-f(x) = f'(c)(x+1-x)$ where c belongs to $\displaystyle (x,x+1)$. Hence, $\displaystyle f(x+1)-f(x) = f'(c)$.
When x goes to infinity, x+1 goes to infinity and so does c. So, both f(x) and f(x+1) go to the limit L. Therefore, f'(c) goes to L-L = 0.
Hope this helps

5. thank you. this one makes more sense and now i see where you are coming from with the MVT.

6. You can also use L'Hospital's Rule; since both those limits are finite we can write $\displaystyle \lim_{x\to \infty}[f(x)+f'(x)] = L$. Now let $\displaystyle a(x) = f(x)e^{x}$ and let $\displaystyle b(x)= e^{x}$; observe that $\displaystyle \frac{a(x)}{b(x)} = f(x)$ since $\displaystyle e^{x} \neq 0$ for all $\displaystyle x\in \mathbb{R}$. Since $\displaystyle \lim_{x\to \infty}\frac{a'(x)}{b'(x)} = L$ and $\displaystyle \lim_{x\to \infty}|b(x)| = +\infty$, we can apply L'Hospital's Rule, and so $\displaystyle \lim_{x\to \infty}\frac{a(x)}{b(x)} = \lim_{x\to \infty}f(x) = L$. It then follows that $\displaystyle \lim_{x\to \infty}f'(x) = 0$.