Bài 3 trang 132 SGK Đại số và Giải tích 11

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:

\(\underset{x\rightarrow -3}{\lim}\) \(\dfrac{x^{2 }-1}{x+1}\) \( = \dfrac{{\mathop {\lim }\limits_{x \to  - 3} \left( {{x^2} - 1} \right)}}{{\mathop {\lim }\limits_{x \to  - 3} \left( {x + 1} \right)}} \) \(= \dfrac{{\mathop {\lim }\limits_{x \to  - 3} {x^2} - \mathop {\lim }\limits_{x \to  - 3} 1}}{{\mathop {\lim }\limits_{x \to  - 3} x + \mathop {\lim }\limits_{x \to  - 3} 1}}\) = \(\dfrac{(-3)^{2}-1}{-3 +1} = -4\).

Quảng cáo

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:

\(\underset{x\rightarrow 6}{\lim}\) \(\dfrac{\sqrt{x + 3}-3}{x-6}\) = \(\underset{x\rightarrow 6}{\lim}\dfrac{(\sqrt{x + 3}-3)(\sqrt{x + 3}+3 )}{(x-6) (\sqrt{x + 3}+3 )}\) 
= \(\underset{x\rightarrow 6}{\lim}\) \(\dfrac{x +3-9}{(x-6) (\sqrt{x + 3}+3 )}\) \( = \mathop {\lim }\limits_{x \to 6} \dfrac{{x - 6}}{{\left( {x - 6} \right)\left( {\sqrt {x + 3}  + 3} \right)}} \) \(= \mathop {\lim }\limits_{x \to 6} \dfrac{1}{{\sqrt {x + 3}  + 3}} \) \(= \dfrac{1}{{\mathop {\lim }\limits_{x \to 6} \left( {\sqrt {x + 3}  + 3} \right)}} \) \(= \dfrac{1}{{\mathop {\lim }\limits_{x \to 6} \left( {\sqrt {x + 3} } \right) + 3}} \) \(= \dfrac{1}{{\sqrt {6 + 3}  + 3}}\)= \(\dfrac{1}{6}\).

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