# Finding the limit $\lim\limits_{x\to 0}\;\frac{f(x)}{\sin x}$ if $f(0) = 0$, $f'(0) = k > 0$ and $0 < f''(x) < f(x)$ for $x \in (0, \pi)$

I would appreciate if somebody could help me with the following problem:

$f(x)$ satisfies the conditions $1,2,3$:

1. $f(0)=0$

2. $f'(0)=k>0$

3. $0<x<\pi$ $\Rightarrow$ $0\leq f''(x)\leq f(x)$

Find $\lim\limits_{x\to 0}\;\dfrac{f(x)}{\sin x}$

-
k is a constant you say? –  ABC Mar 21 '13 at 9:22
if it is so, $f''(x) =0$; and so 3rd condition is redundant. –  ABC Mar 21 '13 at 9:23

You can find limit by using L'Hospital's rule $$\lim_{x\rightarrow 0} \frac {f(x)}{\sin x} = \lim_{x\rightarrow 0} \frac {f'(x)}{\cos x} = f'(0) = k$$

-
What about the third condition for $f$? –  Babak S. Mar 17 '13 at 9:35
It isn't necessary to find limit itself. Probably there are more things to find that OP didn't put in his post. –  Kaster Mar 17 '13 at 23:46