# Taylor polynomial

I need your help to solve this question. I tried something, but i can't finish my proof.

Let $f(x)$ be a differentiable function in $(0, \infty )$, so that $|f'(x)|$ is bounded there. Prove that $\lim_{x \to \infty }\frac{f(x)}{x^{2}}=0$.

My solution:

$f(x)=f(x_{0})+f'(c)(x-x_{0})$ for $x_{0}>0$. $|f'(x)|$ is bounded by N: $\left | \frac{f(x)}{x^2} \right | =\left |\frac{f(x_{0})+f'(c)(x-x_{0})}{x^2} \right |\leqslant \left |\frac{f(x_{0})}{x^2} \right |+\left |\frac{f'(c)(x-x_{0})}{x^2} \right |\leqslant \left |\frac{f(x_{0})}{x^2} \right |+N\cdot \left |\frac{x-x_{0}}{x^2} \right |$.

What is the next step? Thanks!

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Well, what's the limit of the right hand side? (You should indicate that $c$ depends on $x$.) – David Mitra Feb 19 '13 at 14:07
@ David Mitra.Oops, the limit of RHS is 0. thank you! – Panka Feb 19 '13 at 15:23

I think taylor is totally the wrong way, because of it's only an approximation. I would use that $|f'|$ bounded implies $f$ is lipschitz continuous with $$L=\sup |f'(x)|$$
I will do this by saying we have a function value on $0$. In fact there always exist a function with a greater domain which is continuous in $[0,\infty)$. So we get $$\lim_{x\rightarrow \infty} \frac{f(x)-f(0)}{x\cdot x}\leq \lim_{x\rightarrow \infty} \left| \frac{|f(x)-f(0)|}{x}\right| \cdot \frac{1}{x} \leq\lim_{x\rightarrow \infty} L\cdot \frac{1}{x}=0$$