1

Let $f$ be a differentiable function in $[1,\infty) \to \mathbb R$ with a continuous derivative for which $$ f'(x) = \frac 1{x+2010f(x)^2} , \quad x\ge 1 $$ and $f(1)=0$. Find $\lim_{x\to\infty} f(x)/x$.

martini
  • 84,101
Plom
  • 661

1 Answers1

1

You can use l'Hôpital. Or: First observe that $\lim_{x\to\infty}f'(x)=0$. Hence for $x_0$, we have for all $x>x_0$ by the IVT that $f(x)-f(x_0)=(x-x_0)f'(\xi)$ with $x_0<\xi<x$. If we choose $x_0$ big enough, we can ensure $|f'(\xi)|<\frac\epsilon2$, hence in $$ \frac {f(x)}x\cdot \left(1+\frac{x_0}{x-x_0}\right)=\frac{f(x)}{x-x_0}=\frac{f(x)-f(x_0)}{x-x_0}+\frac{f(x_0)}{x-x_0}$$ the right hand side is $<\epsilon$ for $x$ big enough. At the same time the factor in parenteses on the left tends to $2$ as $x\to\infty$, hence is $>1$ for $x$ big enough and thus finally $\left|\frac{f(x)}x\right|<\epsilon$.

Hagen von Eitzen
  • 374,180