Giải bài 5 trang 133 – SGK môn Đại số và Giải tích lớp 11

a) 

Từ hình vẽ, ta có:

$\lim\limits_{x\to -\infty }\,f(x)=0,\,\lim\limits_{x\to {{3}^{-}}}\,=-\infty ,\,\lim\limits_{x\to -{{3}^{+}}}\,=+\infty$
b)
$\bullet \lim\limits_{x\to -\infty }\,f(x)=\lim\limits_{x\to -\infty }\,\dfrac{x+2}{{{x}^{2}}-9}\\=\lim\limits_{x\to -\infty }\,\dfrac{{{x}^{2}}\left( \dfrac{1}{x}+\dfrac{2}{{{x}^{2}}} \right)}{{{x}^{2}}\left( 1-\dfrac{9}{{{x}^{2}}} \right)}=\lim\limits_{x\to -\infty }\,\dfrac{\dfrac{1}{x}+\dfrac{2}{{{x}^{2}}}}{1-\dfrac{9}{{{x}^{2}}}}=0$
Vì $\lim\limits_{x\to -\infty }\,\left( \dfrac{1}{x}+\dfrac{2}{{{x}^{2}}} \right)=0$ và $\lim\limits_{x\to -\infty }\,\left( 1-\dfrac{9}{{{x}^{2}}} \right)=1$  
$\bullet \lim\limits_{x\to {{3}^{-}}}\,f(x)=\lim\limits_{x\to {{3}^{-}}}\,\dfrac{x+2}{(x-3)(x+3)}=-\infty$ 
Vì $\lim\limits_{x\to {{3}^{-}}}\,\left( x+2 \right)=5$ và $\lim\limits_{x\to {{3}^{-}}}\,\left( x-3 \right)\left( x+3 \right)=0, x-3<0$ với mọi $x<3$
$\bullet \,\lim\limits_{x\to -{{3}^{+}}}\,f(x)=\lim\limits_{x\to -{{3}^{+}}}\,\dfrac{x+2}{(x-3)(x+3)}=+\infty$

Vì $\lim\limits_{x\to {{-3}^{+}}}\,\left( x+2 \right)=-1$ và$\lim\limits_{x\to {{-3}^{+}}}\,\left( x-3 \right)=-9$; $\lim\limits_{x\to {{-3}^{+}}}\,\left( x+3 \right)=0, x+3>0$ với mọi $x>-3$