Giải bài 5 trang 142 - SGK Đại số và Giải tích lớp 11
Tìm các giới hạn sau:
\(a) \lim\limits_{x\to 2}\,\dfrac{x+3}{{{x}^{2}}+x+4}\)
\(b) \lim\limits_{x\to -3}\,\dfrac{{{x}^{2}}+5x+6}{{{x}^{2}}+3x}\)
\(c) \lim\limits_{x\to {{4}^{-}}}\,\dfrac{2x-5}{x-4}\)
\(d) \lim\limits_{x\to +\infty }\,\left( -{{x}^{3}}+{{x}^{2}}-2x+1 \right)\)
\(e) \lim\limits_{x\to -\infty }\,\dfrac{x+3}{3x-1}\)
\(f) \lim\limits_{x\to -\infty }\,\dfrac{\sqrt{{{x}^{2}}-2x+4}-x}{3x-1}\)
a) \(\lim\limits_{x\to 2}\,\dfrac{x+3}{{{x}^{2}}+x+4}=\dfrac{2+3}{{{2}^{2}}+2+4}=\dfrac{5}{10}=\dfrac{1}{2}\)
b) \(\lim\limits_{x\to -3}\,\dfrac{{{x}^{2}}+5x+6}{{{x}^{2}}+3x}=\lim\limits_{x\to -3}\,\dfrac{\left( x+2 \right)\left( x+3 \right)}{x\left( x+3 \right)}=\lim\limits_{x\to -3}\,\dfrac{x+2}{x}=\dfrac{-3+2}{-3}=\dfrac{1}{3}\)
c) \(\lim\limits_{x\to {{4}^{-}}}\,\dfrac{2x-5}{x-4}\)
Ta có: \( \lim\limits_{x\to {{4}^{-}}}\,\left( x-4 \right)=0,\,\,x-4<0\,\,\forall \,x<4\) và \(\lim\limits_{x\to {{4}^{-}}}\,\left( 2x-5 \right)=8-5=3>0\)
nên \(\lim\limits_{x\to {{4}^{-}}}\,\dfrac{2x-5}{x-4}=-\infty\)
d) \(\lim\limits_{x\to +\infty }\,\left( -{{x}^{3}}+{{x}^{2}}-2x+1 \right)=\lim\limits_{x\to +\infty }\,{{x}^{3}}\left( -1+\dfrac{1}{x}-\dfrac{2}{{{x}^{2}}}+\dfrac{1}{{{x}^{3}}} \right)=-\infty \)
Vì \(\lim\limits_{x\to +\infty }\,{{x}^{3}}=+\infty\) và \(\lim\limits_{x\to +\infty }\,\left( -1+\dfrac{1}{x}-\dfrac{2}{{{x}^{2}}}+\dfrac{1}{{{x}^{3}}} \right)=-1<0\)
e) \(\lim\limits_{x\to -\infty }\,\dfrac{x+3}{3x-1}=\lim\limits_{x\to -\infty }\,\dfrac{1+\dfrac{3}{x}}{3-\dfrac{1}{x}}=\dfrac{1}{3}\)
Vì \(\lim\limits_{x\to -\infty }\,\left( 1+\dfrac{3}{x} \right)=1\) và \(\lim\limits_{x\to -\infty }\,\left( 3-\dfrac{1}{x} \right)=3\)
f)
\(\lim\limits_{x\to -\infty }\,\dfrac{\sqrt{{{x}^{2}}-2x+4}-x}{3x-1}=\lim\limits_{x\to -\infty }\,\dfrac{-x \sqrt{1-\dfrac{2}{x}+\dfrac{4}{{{x}^{2}}}}-x}{x\left( 3-\dfrac{1}{x} \right)}\\=\lim\limits_{x\to -\infty }\,\dfrac{-\sqrt{1-\dfrac{2}{x}+\dfrac{4}{{{x}^{2}}}}-1}{3-\dfrac{1}{x}}=\dfrac{-2}{3}\)
Vì \(x\to -\infty, x < 0, \lim\limits_{x\to -\infty }\,\left( -\sqrt{1-\dfrac{2}{x}+\dfrac{4}{{{x}^{2}}}}-1 \right)=-2\) và \(\lim\limits_{x\to -\infty }\,\left( 3-\dfrac{1}{x} \right)=3\)