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

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k is a constant you say? –  Mr.ØØ7 Mar 21 '13 at 9:22
if it is so, $f''(x) =0$; and so 3rd condition is redundant. –  Mr.ØØ7 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 $f$? –  B. S. Mar 17 '13 at 9:35