Trắc nghiệm Giới hạn của hàm số lớp 11 (có đúng sai, trả lời ngắn)

Ta có \[\mathop {\lim }\limits_{{\rm{x}} \to \pm \infty } {\rm{c}} = {\rm{c}}\].

\[\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{x}}_0}} \left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{ + g}}\left( {\rm{x}} \right)} \right] = {\rm{L + M}}\].

Ta có \[\mathop {\lim }\limits_{{\rm{x}} \to 2} \left( {2 + {\rm{x}}} \right) = 2 + 2 = 4\].

\[\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{x}}_0}} \left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{ + 2g}}\left( {\rm{x}} \right)} \right]{\rm{ = }}\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{x}}_0}} {\rm{f}}\left( {\rm{x}} \right) + \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{x}}_0}} {\rm{2g}}\left( {\rm{x}} \right) = 1 + 2.4 = 9\].

Khi  \[{\rm{x}} \to {3^ + }\]thì  x > 3

Nên  \[{\rm{x}} - 3 > 0 \Rightarrow \left| {{\rm{x}} - 3} \right| = {\rm{x}} - 3\]

\[\mathop {\lim }\limits_{{\rm{x}} \to {3^ + }} \frac{{\left| {{\rm{x}} - 3} \right|}}{{3{\rm{x}} - 9}} = \mathop {\lim }\limits_{{\rm{x}} \to {3^ + }} \frac{{{\rm{x}} - 3}}{{3{\rm{x}} - 9}} = \mathop {\lim }\limits_{{\rm{x}} \to {3^ + }} \frac{{{\rm{x}} - 3}}{{3.\left( {{\rm{x}} - 3} \right)}} = \frac{1}{3}\].

Ta có 3 – x > 0  với mọi x < 3

\[ \Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to {3^ - }} \frac{{3 - {\rm{x}}}}{{\sqrt {27 - {{\rm{x}}^3}} }} = \mathop {\lim }\limits_{{\rm{x}} \to {3^ - }} \frac{{3 - {\rm{x}}}}{{\sqrt {\left( {3 - {\rm{x}}} \right)\left( {9 + 3{\rm{x}} + {{\rm{x}}^2}} \right)} }} = \mathop {\lim }\limits_{{\rm{x}} \to {3^ - }} \frac{{\sqrt {3 - {\rm{x}}} }}{{\sqrt {9 + 3{\rm{x}} + {{\rm{x}}^2}} }} = \frac{{\sqrt {3 - 3} }}{{\sqrt {9 + 3.3 + {3^2}} }} = 0\].

\[\mathop {\lim }\limits_{{\rm{x}} \to \infty } \frac{{{{\rm{x}}^3} + 3{{\rm{x}}^2} + 4}}{{2{{\rm{x}}^3}}} = \mathop {\lim }\limits_{{\rm{x}} \to \infty } \frac{{{{\frac{{\rm{x}}}{{{{\rm{x}}^{\rm{3}}}}}}^3} + {{\frac{{\rm{x}}}{{{{\rm{x}}^{\rm{3}}}}}}^2} + \frac{{\rm{4}}}{{{{\rm{x}}^{\rm{3}}}}}}}{{\frac{{{\rm{2}}{{\rm{x}}^{\rm{3}}}}}{{{{\rm{x}}^{\rm{3}}}}}}} = \mathop {\lim }\limits_{{\rm{x}} \to \infty } \frac{{1 + \frac{1}{{\rm{x}}} + \frac{4}{{{{\rm{x}}^3}}}}}{2} = \frac{1}{2}\].

\[\mathop {\lim }\limits_{{\rm{x}} \to + \infty } \left( {\sqrt {{\rm{x}} + 5} - \sqrt {{\rm{x}} - 6} } \right) = \mathop {\lim }\limits_{{\rm{x}} \to + \infty } \frac{{{\rm{x}} + 5 - {\rm{x}} + 6}}{{\sqrt {{\rm{x}} + 5} + \sqrt {{\rm{x}} - 6} }} = \mathop {\lim }\limits_{{\rm{x}} \to + \infty } \frac{{11}}{{\sqrt {{\rm{x}} + 5} + \sqrt {{\rm{x}} - 6} }} = \frac{{11}}{\infty } = 0\].