a) \(\underset{x\rightarrow -3} {lim} \frac{x^{2 }-1}{x+1}\)xác định trên \(\mathbb{R}\setminus \) {-1}

\(\underset{x\rightarrow -3} {lim} \frac{x^{2 }-1}{x+1}\)= \(\frac{(-3)^{2}-1}{-3 +1} =\frac{9-1}{-2}= -4\).

b) \(\underset{x\rightarrow -2}{lim}\) \(\frac{4-x^{2}}{x + 2}\)xác định trên\(\mathbb{R}\setminus \) {-2}

\(\underset{x\rightarrow -2} {lim} \frac{4-x^{2}}{x + 2}\) = \(\underset{x\rightarrow -2} {lim} \frac{ (2-x)(2+x)}{x + 2} = \underset{x\rightarrow -2}{lim} (2-x) = 4\).

c) \(\underset{x\rightarrow 6}{lim}\) \(\frac{\sqrt{x + 3}-3}{x-6}\)

\(=\underset{x\rightarrow 6}{lim}\) \(\frac{(\sqrt{x + 3}-3)(\sqrt{x + 3}+3 )}{(x-6) (\sqrt{x + 3}+3 )}\)

\(=\underset{x\rightarrow 6}{lim}\) \(\frac{x +3-9}{(x-6) (\sqrt{x + 3}+3 )}\)

\(=\underset{x\rightarrow 6}{lim}\) \(\frac{1}{\sqrt{x+3}+3}\)

\(=\frac{1}{\sqrt{6+3}+3}=\frac{1}{9+3}=\frac{1}{6}\).

d)  \(\underset{x\rightarrow +\infty }{lim} \frac{2x-6}{4-x}\)xác định trên \(\mathbb{R}\setminus \) {4}

\(\underset{x\rightarrow +\infty }{lim} \frac{2x-6}{4-x}=\underset{x\rightarrow +\infty }{lim} \frac{2-\frac{6}{x}}{\frac{4}{x}-1}= \frac{2}{-1}=-2\).

e) \(\underset{x\rightarrow +\infty }{lim} \frac{17}{x^{2}+1}\) xác định trên \(\mathbb{R}\)

Ta có \(\underset{x\rightarrow +\infty }{lim} (x^2+ 1) =\) \(\underset{x\rightarrow +\infty }{lim} x^2\left ( 1 + \frac{1}{x^{2}} \right )  = +\infty \).

Vậy \(\underset{x\rightarrow +\infty }{lim} \frac{17}{x^{2}+1}=0\)

f) \(\underset{x\rightarrow +\infty }{lim} \frac{-2x^{2}+x -1}{3 +x}\)xác định trên \(\mathbb{R}\setminus \) {-3}

Ta có \(\frac{3}{x^{2}}+\frac{1}{x} > 0\) với \(∀x>0\).

 \(\Rightarrow \underset{x\rightarrow +\infty }{lim} \frac{-2x^{2}+x -1}{3 +x}=\underset{x\rightarrow +\infty }{lim} \frac{-2+\frac{1}{x} -\frac{1}{x^{2}}}{\frac{3}{x^{2}} +\frac{1}{x}} = -\infty \),