Lanczos derivative of $f(x)=|x|$ at $x = 0$ 
Consider a function $f:\mathbb{R} \to \mathbb{R}$. The Lanczos derivative, denoted by $DLf$, is a real function defined at each point by the limit $DLf(x_0) =\displaystyle \lim_{h\to 0}\dfrac{3}{2h^3}\int_{-h}^{h}tf(x_0+t)dt$.
Find the Lanczos derivative of $|x|$ at $x_0=0$.

Here are my calculations:
$DLf(x_0)=\displaystyle \lim_{h\to 0}\dfrac{3}{2h^3}\int_{-h}^{h}t|t|dt=\lim_{h\to 0}\dfrac{3}{2h^3} \left(\int_{-h}^{0}-t^2dt+\int_{0}^{h}t^2dt\right)=\lim_{h\to 0}\dfrac{3}{2h^3} \left(-\dfrac{h^3}{3}+\dfrac{h^3}{3}\right)=\lim_{h \to 0} 0=0$
But Wolfram says it is not determined. Are my calculations correct?

 A: I don't see anything wrong with what you did.  Note $g(t) = t\left|t\right|$ is an odd function, so its integral from $-h$ to $h$, for any $h$, even $h \lt 0$, would automatically be $0$. I'm not sure what Wolfram's software is doing, but it's giving you an incorrect result.
Since I'm confident Wolfram wants their software to be as powerful & robust as they can reasonably make it, you may wish to let them know about this, such as by using their contact form.
Update:
Note I just tried it using brackets around the entire expression you're taking a limit of, with the actual text being:

lim ((3/(2x^3))(int from -x to x of t|t|dt)) as x->0

Wolfram Alpha now shows the following:

Although its rendered input picture looks the same as what you got, I believe if you didn't include the extra brackets, its parser was misinterpreting how to calculate the result (e.g., perhaps it was applying the limit only to the fraction of $\frac{3}{2x^3}$). This could help explain why it failed for you, but if there was ambiguity or anything else like that, it should have at least given some sort of warning or other notice to you about this.
Regardless, this shows you should be careful to use extra brackets to help ensure Wolfram's software works as you expect it should. Also, if you choose to send a notice to Wolfram about this, you should probably also say how you can get it to work properly, such as I've shown here.
