Bài tập 23 trang 152 SGK Toán 11 NC

a)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 2} \left( {3{x^2} + 7x + 11} \right)\\
 = \mathop {\lim }\limits_{x \to 2} 3{x^2} + \mathop {\lim }\limits_{x \to 2} 7x + \mathop {\lim }\limits_{x \to 2} 11\\
 = {3.2^2} + 7.2 + 11 = 37
\end{array}\)

b)

\(\mathop {\lim }\limits_{x \to 1} \frac{{x - {x^3}}}{{\left( {2x - 1} \right)\left( {{x^4} - 3} \right)}} = \frac{0}{{ - 2}} = 0\)

c)

\(\mathop {\lim }\limits_{x \to 0} x\left( {1 - \frac{1}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {x - 1} \right) =  - 1\)

d)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 9} \frac{{\sqrt x  - 3}}{{9x - {x^2}}} = \mathop {\lim }\limits_{x \to 9} \frac{{\sqrt x  - 3}}{{ - x\left( {x - 9} \right)}}\\
 = \mathop {\lim }\limits_{x \to 9} \frac{1}{{ - x\left( {\sqrt x  + 3} \right)}} =  - \frac{1}{{54}}
\end{array}\)

e) \(\mathop {\lim }\limits_{x \to \sqrt 3 } \left| {{x^2} - 4} \right| = 1\)

f)

\(\mathop {\lim }\limits_{x \to 2} \sqrt {\frac{{{x^4} + 3x - 1}}{{2{x^2} - 1}}}  = \sqrt {\frac{{{2^4} + 3.2 - 1}}{{{{2.2}^2} - 1}}}  = \sqrt 3 \)

-- Mod Toán 11 HỌC247