16. We could write
[tex]\dfrac{2x+3}{x-3}=\dfrac{2(x-3)+9}{x-3}=2+\dfrac9{x-3}[/tex]
(basically find the quotient/remainder upon dividing [tex]2x+3[/tex] by [tex]x-3[/tex]). Then as [tex]x\to3[/tex], [tex]\dfrac9{x-3}[/tex] diverges to [tex]-\infty[/tex] from the left, and to [tex]+\infty[/tex] from the right. So the limit does not exist.
20. Factorize the numerator:
[tex]\dfrac{x^2+x-6}{x-2}=\dfrac{(x-2)(x+3)}{x-2}[/tex]
We're considering the limit as [tex]x\to2[/tex], which also means that it's not the case that [tex]x=2[/tex]. Because of this, we can cancel the factor of [tex]x-2[/tex] in both numerator and denominator:
[tex]\displaystyle\lim_{x\to2}\frac{x^2+x-6}{x-2}=\lim_{x\to2}x+3[/tex]
[tex]x+3[/tex] exists for all values of [tex]x[/tex] (i.e. it's continuous on its domain), so the limit is the value of [tex]x+3[/tex] at [tex]x=2[/tex], so
[tex]\displaystyle\lim_{x\to2}x+3=2+3=5[/tex]
