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:

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

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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$

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