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

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

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

D.\[\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{x}}_0}} \left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{ + g}}\left( {\rm{x}} \right)} \right] = {\rm{L}} + 3{\rm{M}}\].

Lời giải

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

Câu 3/20

Tính giá trị của giới hạn \[\mathop {\lim }\limits_{{\rm{x}} \to 2} \left( {2 + {\rm{x}}} \right)\]

A. 2.

B. 3.

C. 4.

D. 5.

Lời giải

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

Câu 4/20

Cho các giới hạn \[\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{x}}_0}} {\rm{f}}\left( {\rm{x}} \right){\rm{ = 1}},\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{x}}_0}} {\rm{g}}\left( {\rm{x}} \right){\rm{ = 4}}\].Tính\[\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{x}}_0}} \left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{ + 2g}}\left( {\rm{x}} \right)} \right]\]

A. 4.

B. 8.

C. 9.

D. 10.

Lời giải

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

Câu 5/20

Trong các mệnh đề sau đây mệnh đề nào là đúng?

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

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

C.\[\mathop {\lim }\limits_{{\rm{x}} \to {3^ + }} \frac{{\left| {{\rm{x}} - 3} \right|}}{{3{\rm{x}} - 9}} = 0\].

D. Không tồn tại \[\mathop {\lim }\limits_{{\rm{x}} \to {3^ + }} \frac{{\left| {{\rm{x}} - 3} \right|}}{{3{\rm{x}} - 9}} = 0\].

Lời giải

Khi \[{\rm{x}} \to {3^ + }\]thì x > 3

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

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

Câu 6/20

Giá trị của giới hạn \[\mathop {\lim }\limits_{{\rm{x}} \to {3^ - }} \frac{{3 - {\rm{x}}}}{{\sqrt {27 - {{\rm{x}}^3}} }}\]bằng:

A. 34.

B. 0.

C. 35.

D. 53.

Lời giải

Ta có 3 – x > 0 với mọi x < 3

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

Câu 7/20

Tính giới hạn của hàm số \[\mathop {\lim }\limits_{{\rm{x}} \to \infty } \frac{{{{\rm{x}}^3} + 3{{\rm{x}}^2} + 4}}{{2{{\rm{x}}^3}}}\]

A. 1.

B. 2.

C.\[\frac{2}{3}\].

D.\(\frac{1}{2}\).

Lời giải

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

Câu 8/20

Giá trị của giới hạn \[\mathop {\lim }\limits_{{\rm{x}} \to + \infty } \left( {\sqrt {{\rm{x}} + 5} - \sqrt {{\rm{x}} - 6} } \right)\] là

A. 1.

B. 2.

C. 3.

D. 0.

Lời giải

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

Câu 9/20