# Prove Using L'Hopital's Rule And Mean Value Theorem.

how I can prove, for a function, $$f(x)$$ is differentiable such that $$\lim_{x\rightarrow \infty}f(x)=L\neq 0$$ and $$\lim_{x\rightarrow \infty}f'(x)$$ exists. Prove so that $$\lim_{x\rightarrow\infty}f'(x)=0$$. I have to use L'Hopital's Rule, and The Mean Value Theorem, do it 2 times.

1. How to prove using L'Hopital's Rule. I want to convert to indeterminate form. But I don't know the function, it is a random function. I converted this to $$\frac{e^{x}f(x)}{e^x}$$. What to do next?

2. How to prove using The Mean Value Theorem. I think I can use Mean Value Theorem to $$f(x)$$ on the interval (x, x+1), where x can be any number. And then calculate to $$x\rightarrow\infty$$. How to do that?

I have to prove this both ways.

Sketch: 1.$$\lim_{x\to\infty}\frac{e^x f(x)}{e^x}=\lim_{x\to\infty}(f’(x)+f(x))\,,$$ which implies $$L=\lim_{x\to\infty}f’(x)+L\,.$$
2.$$f(x+h)-f(x)=f’(\xi)h \,,$$ yields $$L-L=\lim_{x\to\infty}f’(x) h\,.$$
As shown in Baby Rudin we can apply L'Hopital's Rule as long as the denominator tends to $$\infty$$; it is not necessary that we have $$\frac {\infty} {\infty}$$ form. So apply the rule to $$\lim \frac {f(x)} x$$ to show that $$f'(x) \to 0$$.
Hint for proof using MVT: If $$f'(x) \ to l >0$$ the $$f(n+1)-f(n) \geq \frac l 2$$ for all $$n$$ sufficiently large. Show using this that $$(f(n)$$ is not even bounded.