a) Quan sát đồ thị ta thấy \(x → -∞\) thì \(f(x) → 0\)

Khi \(x → 3^-\) thì \(f(x) → -∞\);

Khi \(x → -3^+\) thì \(f(x)  → +∞\).

b) \(\underset{x\rightarrow -\infty }{lim} f(x) = \underset{x\rightarrow -\infty }{lim} \frac{x+2}{x^{2}-9}=\underset{x\rightarrow -\infty }{lim} \frac{\frac{1}{x}+\frac{2}{x^{2}}}{1-\frac{9}{x^{2}}} = 0\).

Vì  \(\underset{x\rightarrow 3^{-}}{lim} \frac{x+2}{x+3}=\frac{3+2}{3+3}=\frac{5}{6} > 0\) và \(\underset{x\rightarrow 3^{-}}{\lim} \frac{1}{x-3} = -∞\).

\(\Rightarrow \underset{x\rightarrow 3^{-}}{lim} f(x) = \underset{x\rightarrow 3^{-}}{lim} \frac{x+2}{x^{2}-9} =  \underset{x\rightarrow 3^{-}}{lim} \frac{x+2}{x+3}.\frac{1}{x-3} = -∞ \)

Vì  \(\underset{x\rightarrow -3^{+}}{lim} \frac{x+2}{x-3}=\frac{-3+2}{-3-3}= \frac{-1}{-6} = \frac{1}{6} > 0\) và \(\underset{x\rightarrow -3^{+}}{lim} \frac{1}{x+3} = +∞\).

\(\Rightarrow \underset{x\rightarrow -3^{+}}{lim} f(x) =\underset{x\rightarrow -3^{+}}{lim} \frac{x+2}{x^{2}-9} = \underset{x\rightarrow -3^{+}}{lim} \frac{x+2}{x-3} . \frac{1}{x+3} = +∞\)