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Bài tập 38 trang 166 SGK Toán 11 NC

Bài tập 38 trang 166 SGK Toán 11 NC

Tìm các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 8}}{{{x^2} - 4}}\)

b) \(\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ + }} \frac{{2{x^2} + 5x - 3}}{{{{\left( {x + 3} \right)}^2}}}\)

c) \(\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} \frac{{2{x^2} + 5x - 3}}{{{{\left( {x + 3} \right)}^2}}}\)

d) \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^3} + 1}  - 1}}{{{x^2} + x}}\)

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Hướng dẫn giải chi tiết

 
 

a)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 8}}{{{x^2} - 4}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + 2x + 4}}{{x + 2}} = 3
\end{array}\)

b)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ + }} \frac{{2{x^2} + 5x - 3}}{{{{\left( {x + 3} \right)}^2}}} = \mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ + }} \frac{{\left( {x + 3} \right)\left( {2x - 1} \right)}}{{{{\left( {x + 3} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ + }} \frac{{2x - 1}}{{x + 3}} =  - \infty 
\end{array}\)

vì \(\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ + }} \left( {2x - 1} \right) =  - 7 < 0,\)

\(\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ + }} \left( {x + 3} \right) = 0\) và x + 3 > 0.

c)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} \frac{{2{x^2} + 5x - 3}}{{{{\left( {x + 3} \right)}^2}}} = \mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} \frac{{\left( {x + 3} \right)\left( {2x - 1} \right)}}{{{{\left( {x + 3} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} \frac{{2x - 1}}{{x + 3}} =  + \infty 
\end{array}\)

(vì \(\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} \left( {2x - 1} \right) =  - 7 < 0,\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} \left( {x + 3} \right) = 0\) và x - 3 < 0)

d)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^3} + 1}  - 1}}{{{x^2} + x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{x^3}}}{{x\left( {x + 1} \right)\left( {\sqrt {{x^3} + 1}  + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}}}{{\left( {x + 1} \right)\left( {\sqrt {{x^3} + 1}  + 1} \right)}} = 0
\end{array}\)

-- Mod Toán 11 HỌC247

 
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