Bài 3 trang 132 SGK Đại số và Giải tích 11

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

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 \(\mathop {\lim }\limits_{x \to  + \infty } \left( {\frac{3}{{{x^2}}} + \frac{1}{x}} \right) = 0\).