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Today I had an exam and the following problem came up. I have absolutely no idea how to approach this. Any help in solving this is appreciated!

$$ \lim_{x\to 0} \frac{\mathrm d^2}{\mathrm dx^2} \frac{f(x)}{x},\qquad f(0) = 0$$

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    Heuristically, viewing $f$ as a Taylor series $$f(x)=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{6}f'''(0)+\cdots,$$ one should expect the answer to be $f'''(0)/3$. I suppose using the quotient rule twice and then doing some clever fractional rearrangement might provide the intended route to the answer, or maybe not.2011-09-06
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    Are there other hypothesis on $f$?2011-09-06
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    Well, actually yes. I will look it up as soon as possible and enhance the question. It was something along the lines of f is in $C^3$ but not sure anymore.2011-09-06
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    Afeter deriving, have you tried l'Hôpital's rule?2011-09-06
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    @anon: Excellent idea, this definitely leads to the solution. However the correct answer IMHO is f′′′(0) * 11/6. Remove the f(0) (which is zero), divide the rest by x, differentiate twice and see which terms are not multiplied by x.2011-09-06
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    @valdo: $\frac{d^2}{dx^2} \frac{1}{6}f'''(0)x^2|_{x=0}=f'''(0)/3$, don't know where you're getting $11/6$ from.2011-09-06
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    @anon: Oh, I understood my mistake. When differentiating the Tailor series with respect to x I also differentiated the f'(0), f''(0) and f'''(0). Which is wrong, since those are just constants.2011-09-06

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The second derivative of $f(x)/x$ can be found with two applications of the quotient rule: $$\frac{x^2f''(x)-2xf'(x)+2f(x)}{x^3}.$$ Now to evaluate the limit of this as $x\to0$ we can take Iasafro's suggestion from the comments of using a trick called L'Hôpital's rule. Taking the derivative of numerator and denominator above leads to a lot of cancelling terms, which comes out to be $$\frac{f'''(x)}{3}.$$ Taking the limit gives $f'''(0)/3$.