Bài 3. Tính các giới hạn sau:
a) \(\underset{x\rightarrow -3}{lim}\) \(\frac{x^{2 }-1}{x+1}\);
b) \(\underset{x\rightarrow -2}{lim}\) \(\frac{4-x^{2}}{x + 2}\);
c) \(\underset{x\rightarrow 6}{lim}\) \(\frac{\sqrt{x + 3}-3}{x-6}\);
d) \(\underset{x\rightarrow +\infty }{lim}\) \(\frac{2x-6}{4-x}\);
e) \(\underset{x\rightarrow +\infty }{lim}\) \(\frac{17}{x^{2}+1}\);
f) \(\underset{x\rightarrow +\infty }{lim}\) \(\frac{-2x^{2}+x -1}{3 +x}\).
Hướng dẫn giải:
a) \(\underset{x\rightarrow -3}{lim}\) \(\frac{x^{2 }-1}{x+1}\) = \(\frac{(-3)^{2}-1}{-3 +1} = -4\).
b) \(\underset{x\rightarrow -2}{lim}\) \(\frac{4-x^{2}}{x + 2}\) = \(\underset{x\rightarrow -2}{lim}\) \(\frac{ (2-x)(2+x)}{x + 2}\) = \(\underset{x\rightarrow -2}{lim} (2-x) = 4\).
c) \(\underset{x\rightarrow 6}{lim}\) \(\frac{\sqrt{x + 3}-3}{x-6}\) = \(\underset{x\rightarrow 6}{lim}\) \(\frac{(\sqrt{x + 3}-3)(\sqrt{x + 3}+3 )}{(x-6) (\sqrt{x + 3}+3 )}\) 
= \(\underset{x\rightarrow 6}{lim}\) \(\frac{x +3-9}{(x-6) (\sqrt{x + 3}+3 )}\) = \(\underset{x\rightarrow 6}{lim}\) \(\frac{1}{\sqrt{x+3}+3}\) = \(\frac{1}{6}\).
d)  \(\underset{x\rightarrow +\infty }{lim}\) \(\frac{2x-6}{4-x}\) =  \(\underset{x\rightarrow +\infty }{lim}\) \(\frac{2-\frac{6}{x}}{\frac{4}{x}-1} = -2\).
e) \(\underset{x\rightarrow +\infty }{lim}\) \(\frac{17}{x^{2}+1} = 0\) vì \(\underset{x\rightarrow +\infty }{lim}\)  \((x^2+ 1) =\) \(\underset{x\rightarrow +\infty }{lim} x^2( 1 + \frac{1}{x^{2}}) = +∞\).
f) \(\underset{x\rightarrow +\infty }{lim}\) \(\frac{-2x^{2}+x -1}{3 +x}\) = \(\underset{x\rightarrow +\infty }{lim}\) \(\frac{-2+\frac{1}{x} -\frac{1}{x^{2}}}{\frac{3}{x^{2}} +\frac{1}{x}} = -∞\), vì \(\frac{3}{x^{2}}+\frac{1}{x} > 0\) với \(∀x>0\).