Question 14
As we approaching from the negative side, [tex]|x-3|=3-x[/tex].
So, we have [tex]\lim_{x \to 3^{-}}=\frac{3-x}{x-3}=\frac{-(x-3)}{x-3}=\boxed{-1}[/tex]
and thus the limit exists at x=3.
Question 15
For the top part of the function, [tex](9-9)^{2}=0[/tex]
For the bottom part of the function, [tex]9-9=0[/tex]
As both one-sided limits are equal, L = 0, and thus the limit exists at x=9.
