# Calculating Limit $\lim\limits_{x\to 0} \frac{\mathrm d^2}{\mathrm dx^2} \frac{f(x)}{x}$

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. – anon Sep 6 '11 at 9:42
Are there other hypothesis on $f$? – Davide Giraudo Sep 6 '11 at 9:44
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. – entrance_exam Sep 6 '11 at 10:00
Afeter deriving, have you tried l'Hôpital's rule? – Josué Tonelli-Cueto Sep 6 '11 at 10:01
@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. – valdo Sep 6 '11 at 10:27

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 $f'''(0)/3$.