a)
$\lim_{x \to 0} \dfrac{4x}{\sqrt{9+x}-3}$
$= \lim_{x \to 0} \dfrac{4x}{\dfrac{(\sqrt{9+x}-3)(\sqrt{9+x}+3)}{\sqrt{9+x}+3}}$
$= \lim_{x \to 0} \dfrac{4x}{\dfrac{9+x-9}{\sqrt{9+x}+3}}$
$= \lim_{x \to 0} \dfrac{4x}{\dfrac{x}{\sqrt{9+x}+3}}$
$= \lim_{x \to 0} 4(\sqrt{9+x}+3)$
$= 4.(\sqrt{9+0}+3) = 24$
b)
$\lim_{x \to 3^{+}} \dfrac{1-2x²}{x-3}$
Ta có: $\lim_{x \to 3^{+}}1-2x² = 1-2.3² = -17 < 0$
Ta lại có: $x > 3$
$⇔x - 3 > 0$
Vậy $\lim_{x \to 3^{+}} \dfrac{1-2x²}{x-3} = -∞$
