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.
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.
$$f(x) = \frac{x^2\cdot\lfloor x\rfloor}{|x-1|}$$
Assume $0 \le \delta &lt; 1$. Then
\begin{align}
   f(3)&amp;= 13.5\\
   f(3-\delta) &amp;= 9 \\
\hline
   f'(3^-) 
   &amp;= \lim_{\delta \to 0^+}\dfrac{f(3) - f(3-\delta)}{3 - (3-\delta)} \\
   &amp;= \lim_{\delta \to 0^+}\dfrac{f(3) - f(3-\delta)}{3 - (3-\delta)} \\
   &amp;= \lim_{\delta \to 0^+}\dfrac{4.5}{\delta} \\
   &amp;= \infty
\end{align}
You did it correctly. Why? Because the floor function is discontinuous from below.
$f(3^-) = 9$ and $f(3) = f(3^+) = 13.5$