Giải bài 13 trang 180 – SGK môn Đại số và Giải tích 11
Tính các giới hạn sau:
\(a)\,\lim\limits_{x\to -2}\,\dfrac{6-3x}{\sqrt{2{{x}^{2}}+1}};\)
\(b)\,\lim\limits_{x\to 2}\,\dfrac{x-\sqrt{3x-2}}{{{x}^{2}}-4};\)
\(c)\,\lim\limits_{x\to {{2}^{+}}}\,\dfrac{{{x}^{2}}-3x+1}{x-2};\)
\(d)\,\lim\limits_{x\to {{1}^{-}}}\,\left( x+{{x}^{2}}+...+{{x}^{n}}-\dfrac{n}{1-x} \right);\)
\(e)\,\lim\limits_{x\to +\infty }\,\dfrac{2x-1}{x+3}; \)
\(f)\,\lim\limits_{x\to -\infty }\,\dfrac{x+\sqrt{4{{x}^{2}}-1}}{2-3x}; \)
\(g)\,\lim\limits_{x\to -\infty }\,\left( -2{{x}^{3}}+{{x}^{2}}-3x+1 \right).\)
Hướng dẫn:
Xem lại định lý về giới hạn hữu hạn và các quy tắc tìm giới hạn vô cực của hàm số
\(\begin{align} & a)\,\lim\limits_{x\to -2}\,\dfrac{6-3x}{\sqrt{2{{x}^{2}}+1}} \\ & =\lim\limits_{x\to -2}\,\dfrac{6-3.(-2)}{\sqrt{2.{{\left( -2 \right)}^{2}}+1}}=\dfrac{12}{3}=4 \\ & b)\,\lim\limits_{x\to 2}\,\dfrac{x-\sqrt{3x-2}}{{{x}^{2}}-4} \\ & =\lim\limits_{x\to 2}\,\dfrac{{{x}^{2}}-3x+2}{\left( x-2 \right)\left( x+2 \right)\left( x+\sqrt{3x-2} \right)} \\ & =\lim\limits_{x\to 2}\,\dfrac{x-1}{\left( x+2 \right)\left( x+\sqrt{3x-2} \right)}=\dfrac{1}{16} \\ & c)\,\lim\limits_{x\to {{2}^{+}}}\,\dfrac{{{x}^{2}}-3x+1}{x-2}=-\infty \\ &\,\,\, \text{vì }\,\lim\limits_{x\to {{2}^{+}}}\,\left( {{x}^{2}}-3x+1 \right)=-1<0 \\ &\,\,\, \text{và}\,\,\lim\limits_{x\to {{2}^{+}}}\,\left( x-2 \right)=0;\,\,x-2>0\,\,\forall x>2 \\ & d)\,\lim\limits_{x\to {{1}^{-}}}\,\left( x+{{x}^{2}}+...+{{x}^{n}}-\dfrac{n}{1-x} \right) \\ & =\lim\limits_{x\to {{1}^{-}}}\,\dfrac{x\left( 1+x+...+{{x}^{n-1}} \right)\left( 1-x \right)-n}{1-x} \\ & =\lim\limits_{x\to {{1}^{-}}}\,\dfrac{x{{\left( 1-x \right)}^{n}}-n}{1-x}=-\infty \\ & e)\,\lim\limits_{x\to +\infty }\,\dfrac{2x-1}{x+3}=\lim\limits_{x\to +\infty }\,\dfrac{2-\dfrac{1}{x}}{1+\dfrac{3}{x}}=2; \\ & f)\,\lim\limits_{x\to -\infty }\,\dfrac{x+\sqrt{4{{x}^{2}}-1}}{2-3x} \\ & =\lim\limits_{x\to -\infty }\,\dfrac{x\left( 1-\sqrt{4-\dfrac{1}{{{x}^{2}}}} \right)}{x\left( \dfrac{2}{x}-3 \right)}=\dfrac{1-2}{-3}=\dfrac{1}{3} \\ & g)\,\lim\limits_{x\to -\infty }\,\left( -2{{x}^{3}}+{{x}^{2}}-3x+1 \right)=+\infty . \\ \end{align}. \)