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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
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1 Answer 1

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

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What about the third condition for $f$? –  B. 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
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