Side-limit of $f(x)=\frac{x^2\cdot\lfloor x\rfloor}{|x-1|}$ The problem wants the value of left hand limit of the derivative of $f(x)=\frac{x^2\cdot\lfloor x\rfloor}{|x-1|}$ when $x\to3^-$.I graphed the function and the answer is $\frac{3}{2}$ but what's wrong with my method:
For any value $x\to3^-$ we have $[x]=2$ , $|x-1|=x-1$ so if $h\to0^-$, then $f(3+h)=\frac{2(3+h)^2}{2+h}=\frac{2h^2+12h+18}{2+h}$. So the value of the derivative is equal to:
$$\lim_{h\to0^-}\dfrac{f(3+h)-f(3)}{h}=\lim_{h\to0^-}\frac{\frac{2h^2+12h+18}{2+h}-\frac{27}{2}}{h}$$ which is infinity and clearly wrong with or without the graph. I'm not new into this but I can't figure out what's wrong here despite I thought about the method for one hour straight.
 A: I would disagree with you that the answer is 3/2. The left hand limit is +inf, and the right hand limit is -inf, as you can see if you graph the function of the derivative.
A: Alright I got the problem. In the following figure the green graph is $f(x)$ and the red one is $f'(x)$. Now if you calculate the left hand limit of the derivative using $\displaystyle\lim_{h\to0^-}\dfrac{f(x+h)-f(x)}{h}$, you're actually calculating the slope of the line that approaches to the segment $AB$ which is infinity so we're not wrong on this point.

The major problem about this method is that we actually want the slope of the line tangent to the graph of $f(x)$ when we're still before $x=3$. For doing that simply note that for $x\in(2,3)$ which contains the left neighborhoods of $x=3$, $f(x)=\frac{2x^2}{x-1}$ and this function's derivative in $x=3$ is exactly equal to $\frac{2}{3}$.
Furthermore, if you insist on solving this with the definition of derivative one can modify the formula like the following: (the following form is clearly a more general definition that will also work for the discontinuity points)
$$\displaystyle {f'_\pm} (x)=\lim_{h\to0^{\pm}}\bigg(\lim_{t\to0^{\pm}}\dfrac{f(x+h+t)-f(x+t)}{h}\bigg)$$
In which we shift the points in the wanted direction first, and then do the rest.
*P.S. The segment $AB$ is not a part of the graph. I added it myself
A: $$f(x) = \frac{x^2\cdot\lfloor x\rfloor}{|x-1|}$$
Assume $0 \le \delta < 1$. Then
\begin{align}
   f(3)&= 13.5\\
   f(3-\delta) &= 9 \\
\hline
   f'(3^-) 
   &= \lim_{\delta \to 0^+}\dfrac{f(3) - f(3-\delta)}{3 - (3-\delta)} \\
   &= \lim_{\delta \to 0^+}\dfrac{f(3) - f(3-\delta)}{3 - (3-\delta)} \\
   &= \lim_{\delta \to 0^+}\dfrac{4.5}{\delta} \\
   &= \infty
\end{align}
You did it correctly. Why? Because the floor function is discontinuous from below.
$f(3^-) = 9$ and $f(3) = f(3^+) = 13.5$
