Bài 3 trang 132 SGK Đại số và Giải tích 11Tính các giới hạn sau: Quảng cáo
Video hướng dẫn giải Tính các giới hạn sau: LG a \(\underset{x\rightarrow -3}{\lim}\) \(\frac{x^{2 }-1}{x+1}\); Phương pháp giải: Nếu hàm số \(y=f(x)\) xác định tại \(x=x_0\) thì \(\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = f\left( {{x_0}} \right)\). Nếu giới hạn hàm số có dạng vô định, tìm cách khử dạng vô định. Lời giải chi tiết: \(\underset{x\rightarrow -3}{\lim}\) \(\dfrac{x^{2 }-1}{x+1}\) \( = \dfrac{{\mathop {\lim }\limits_{x \to - 3} \left( {{x^2} - 1} \right)}}{{\mathop {\lim }\limits_{x \to - 3} \left( {x + 1} \right)}} \) \(= \dfrac{{\mathop {\lim }\limits_{x \to - 3} {x^2} - \mathop {\lim }\limits_{x \to - 3} 1}}{{\mathop {\lim }\limits_{x \to - 3} x + \mathop {\lim }\limits_{x \to - 3} 1}}\) = \(\dfrac{(-3)^{2}-1}{-3 +1} = -4\). LG b \(\underset{x\rightarrow -2}{\lim}\) \(\dfrac{4-x^{2}}{x + 2}\); Lời giải chi tiết: \(\underset{x\rightarrow -2}{\lim}\) \(\dfrac{4-x^{2}}{x + 2}\) = \(\underset{x\rightarrow -2}{\lim}\) \(\dfrac{ (2-x)(2+x)}{x + 2}\) = \(\underset{x\rightarrow -2}{\lim} (2-x) =2-(-2)= 4\) LG c \(\underset{x\rightarrow 6}{\lim}\) \(\dfrac{\sqrt{x + 3}-3}{x-6}\) Lời giải chi tiết: \(\underset{x\rightarrow 6}{\lim}\) \(\dfrac{\sqrt{x + 3}-3}{x-6}\) = \(\underset{x\rightarrow 6}{\lim}\dfrac{(\sqrt{x + 3}-3)(\sqrt{x + 3}+3 )}{(x-6) (\sqrt{x + 3}+3 )}\) LG d \(\underset{x\rightarrow +\infty }{\lim}\) \(\dfrac{2x-6}{4-x}\) Lời giải chi tiết: \(\underset{x\rightarrow +\infty }{\lim}\) \(\dfrac{2x-6}{4-x}\) \( = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{x\left( {2 - \dfrac{6}{x}} \right)}}{{x\left( {\dfrac{4}{x} - 1} \right)}} \) \(= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{2 - \dfrac{6}{x}}}{{\dfrac{4}{x} - 1}} \) \(= \dfrac{{2 - \mathop {\lim }\limits_{x \to + \infty } \dfrac{6}{x}}}{{\mathop {\lim }\limits_{x \to + \infty } \dfrac{4}{x} - 1}} \) \(= \dfrac{{2 - 0}}{{0 - 1}}\) \( = -2\) LG e \(\underset{x\rightarrow +\infty }{\lim}\) \(\dfrac{17}{x^{2}+1}\) Lời giải chi tiết: \(\underset{x\rightarrow +\infty }{\lim}\) \(\dfrac{17}{x^{2}+1} = 0\) vì: \(\underset{x\rightarrow +\infty }{\lim}\) \((x^2+ 1) =\) \(\underset{x\rightarrow +\infty }{\lim} x^2( 1 + \dfrac{1}{x^{2}}) = +∞\) Cách khác: \(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{17}}{{{x^2} + 1}}\) \( = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^2}.\dfrac{{17}}{{{x^2}}}}}{{{x^2}.\left( {1 + \dfrac{1}{{{x^2}}}} \right)}} \) \(= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\dfrac{{17}}{{{x^2}}}}}{{1 + \dfrac{1}{{{x^2}}}}} \) \(= \dfrac{{\mathop {\lim }\limits_{x \to + \infty } \dfrac{{17}}{{{x^2}}}}}{{1 + \mathop {\lim }\limits_{x \to + \infty } \dfrac{1}{{{x^2}}}}} \) \(= \dfrac{0}{{1 + 0}} = 0\) LG f \(\underset{x\rightarrow +\infty }{\lim}\) \(\dfrac{-2x^{2}+x -1}{3 +x}\) Lời giải chi tiết: \(\underset{x\rightarrow +\infty }{\lim}\) \(\dfrac{-2x^{2}+x -1}{3 +x}\) \(= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^2}\left( { - 2 + \dfrac{1}{x} - \dfrac{1}{{{x^2}}}} \right)}}{{{x^2}\left( {\dfrac{3}{{{x^2}}} + \dfrac{1}{x}} \right)}}\) \(=\underset{x\rightarrow +\infty }{\lim}\dfrac{-2+\dfrac{1}{x} -\dfrac{1}{x^{2}}}{\dfrac{3}{x^{2}} +\dfrac{1}{x}} \) Vì \(\mathop {\lim }\limits_{x \to + \infty } \left( {\dfrac{3}{{{x^2}}} + \dfrac{1}{x}} \right) = 0\); \({\dfrac{3}{{{x^2}}} + \dfrac{1}{x}}>0\) khi \(x \to + \infty\) và \(\mathop {\lim }\limits_{x \to + \infty } \left( { - 2 + \dfrac{1}{x} - \dfrac{1}{{{x^2}}}} \right) \) \(= - 2 + \mathop {\lim }\limits_{x \to + \infty } \dfrac{1}{x} - \mathop {\lim }\limits_{x \to + \infty } \dfrac{1}{{{x^2}}}\) \( = - 2 + 0 - 0 = - 2 < 0\) Vậy \(\underset{x\rightarrow +\infty }{\lim}\) \(\dfrac{-2x^{2}+x -1}{3 +x}\)\(=\underset{x\rightarrow +\infty }{\lim}\dfrac{-2+\dfrac{1}{x} -\dfrac{1}{x^{2}}}{\dfrac{3}{x^{2}} +\dfrac{1}{x}} \) \(=-\infty \) Cách khác: \(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{ - 2{x^2} + x - 1}}{{3 + x}} \) \(= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^2}\left( { - 2 + \dfrac{1}{x} - \dfrac{1}{{{x^2}}}} \right)}}{{x\left( {\dfrac{3}{x} + 1} \right)}}\) \( = \mathop {\lim }\limits_{x \to + \infty } \left[ {x.\dfrac{{ - 2 + \dfrac{1}{x} - \dfrac{1}{{{x^2}}}}}{{\dfrac{3}{x} + 1}}} \right]\) Mà \(\mathop {\lim }\limits_{x \to + \infty } x = + \infty \) và \(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{ - 2 + \dfrac{1}{x} - \dfrac{1}{{{x^2}}}}}{{\dfrac{3}{x} + 1}} \) \(= \dfrac{{ - 2 + \mathop {\lim }\limits_{x \to + \infty } \dfrac{1}{x} - \mathop {\lim }\limits_{x \to + \infty } \dfrac{1}{{{x^2}}}}}{{\mathop {\lim }\limits_{x \to + \infty } \dfrac{3}{x} + 1}} \) \(= \dfrac{{ - 2 + 0 - 0}}{{0 + 1}} = - 2 < 0\) Nên \(\underset{x\rightarrow +\infty }{\lim}\) \(\dfrac{-2x^{2}+x -1}{3 +x}\)\(=-\infty \) Loigiaihay.com
Quảng cáo
|