Bài tập 4.55 trang 173 SBT Toán 11

a)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + 1}  - 1}}{{4 - \sqrt {{x^2} + 16} }} = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2}\left( {4 + \sqrt {{x^2} + 16} } \right)}}{{\left( { - {x^2}} \right)\left( {\sqrt {{x^2} + 1}  + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{ - \left( {4 + \sqrt {{x^2} + 16} } \right)}}{{\sqrt {{x^2} + 1}  + 1}} =  - 4
\end{array}\)

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

c) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{2{x^4} + 5x - 1}}{{1 - {x^2} + {x^4}}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{2 + \frac{5}{{{x^3}}} - \frac{1}{{{x^4}}}}}{{\frac{1}{{{x^4}}} - \frac{1}{{{x^2}}} + 1}} = 2\)

d)
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - \infty } \frac{{x + \sqrt {4{x^2} - x + 1} }}{{1 - 2x}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{x + \left| x \right|\sqrt {4 - \frac{1}{x} + \frac{1}{{{x^2}}}} }}{{x\left( {\frac{1}{x} - 2} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{1 - \sqrt {4 - \frac{1}{x} + \frac{1}{{{x^2}}}} }}{{\frac{1}{x} - 2}} = \frac{1}{2}
\end{array}\)

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

f)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {2^ + }} x\left( {\frac{1}{{{x^2} - 4}} - \frac{1}{{x - 2}}} \right) = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{1 - \left( {x + 2} \right)}}{{{x^2} - 4}}\\
 = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{ - x - 1}}{{{x^2} - 4}} =  - \infty 
\end{array}\)

-- Mod Toán 11 HỌC247