Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

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

share|cite|improve this question
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 $f'''(0)/3$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.