Bài tập 41 trang 166 SGK Toán 11 NC

Bài tập 41 trang 166 SGK Toán 11 NC

a)

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

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2x - {x^2}}  - 1}}{{{x^2} - x}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{2x - {x^2} - 1}}{{x\left( {x - 1} \right)\left( {\sqrt {2x - {x^2}}  + 1} \right)}}
\end{array}\\
\begin{array}{l}
 = \mathop {\lim }\limits_{x \to 1} \frac{{ - {{\left( {x - 1} \right)}^2}}}{{x\left( {x - 1} \right)\left( {\sqrt {2x - {x^2}}  + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{1 - x}}{{x\left( {\sqrt {2x - {x^2}}  + 1} \right)}} = 0
\end{array}
\end{array}\)

-- Mod Toán 11