60 Câu Trắc Nghiệm Về Giới Hạn Của Hàm Số Giải Chi Tiết

Chọn A

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

Chọn B

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

Chọn D

Ta có $\mathop {\lim }\limits_{x \to 0} {x^2}\sin \frac{1}{2} = 0.\sin \frac{1}{2} = 0$

Chọn B

$\mathop {\lim }\limits_{x \to – 1} \frac{{{x^2} – 3}}{{{x^3} + 2}} = \frac{{{{\left( { – 1} \right)}^2} – 3}}{{{{\left( { – 1} \right)}^3} + 2}} = – 2$

Chọn C

$\mathop {\lim }\limits_{x \to 1} \frac{{x – {x^3}}}{{\left( {2x – 1} \right)\left( {{x^4} – 3} \right)}} = \frac{{1 – {1^3}}}{{\left( {2.1 – 1} \right)\left( {{1^4} – 3} \right)}} = 0$

Chọn D

Ta có $\mathop {\lim }\limits_{x \to – 1} \,\,\frac{{\left| {x – 1} \right|}}{{{x^4} + x – 3}} = \frac{{\left| { – 1 – 1} \right|}}{{1 – 1 – 3}} = – \frac{2}{3}$

Chọn A

Ta có $\mathop {\lim }\limits_{x \to – 1} \frac{{\sqrt {3{x^2} + 1} – x}}{{x – 1}} = \frac{{\sqrt {3 + 1} + 1}}{{ – 1 – 1}} = – \frac{3}{2}$

Chọn C

$\mathop {\lim }\limits_{x \to 3} \sqrt {\frac{{9{x^2} – x}}{{\left( {2x – 1} \right)\left( {{x^4} – 3} \right)}}} = \sqrt {\frac{{{{9.3}^2} – 3}}{{\left( {2.3 – 1} \right)\left( {{3^4} – 3} \right)}}} = \frac{1}{{\sqrt 5 }}$

Chọn B

$\mathop {\lim }\limits_{x \to 2} \sqrt[3]{{\frac{{{x^2} – x + 1}}{{{x^2} + 2x}}}} = \sqrt {\frac{{{2^2} – 2 + 1}}{{{2^2} + 2.2}}} = \frac{1}{2}$

Chọn C

Ta có: $\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{3{x^2} – 4}} – \sqrt {3x – 2} }}{{x + 1}} = \frac{{\sqrt[3]{{12 – 4}} – \sqrt {6 – 2} }}{3} = \frac{0}{3} = 0$

Chọn D

$\mathop {\lim }\limits_{x \to – \infty } {\mkern 1mu} \left( {x – {x^3} + 1} \right) = \mathop {\lim }\limits_{x \to – \infty } {\mkern 1mu} {x^3}\left( {\frac{1}{{{x^2}}} – 1 + \frac{1}{{{x^3}}}} \right) = + \infty $ vì $\left\{ \begin{gathered}
\mathop {\lim }\limits_{x \to – \infty } {\mkern 1mu} {x^3} = – \infty \hfill \\
\mathop {\lim }\limits_{x \to – \infty } {\mkern 1mu} \left( {\frac{1}{{{x^2}}} – 1 + \frac{1}{{{x^3}}}} \right) = – 1 < 0 \hfill \\
\end{gathered} \right.$

Giải nhanh: $x – {x^3} + 1\sim\left( { – 1} \right){x^3}\xrightarrow[{}]{{}} + \infty $ khi $x \to – \infty .$

Chọn B

Ta có

$\mathop {\lim }\limits_{x \to – \infty } {\mkern 1mu} \left( {{{\left| x \right|}^3} + 2{x^2} + 3\left| x \right|} \right)$

$ = \mathop {\lim }\limits_{x \to – \infty } {\mkern 1mu} \left( { – {x^3} + 2{x^2} – 3x} \right) = \mathop {\lim }\limits_{x \to – \infty } {\mkern 1mu} {x^3}\left( { – 1 + \frac{2}{x} – \frac{3}{{{x^2}}}} \right) = + \infty $

Giải nhanh: ${\left| x \right|^3} + 2{x^2} + 3\left| x \right|\sim{\left| x \right|^3} \to + \infty $ khi $x \to – \infty .$

Chọn B

Giải nhanh: $x \to + \infty :{\mkern 1mu} {\mkern 1mu} \sqrt {{x^2} + 1} + x\sim\sqrt {{x^2}} + x = 2x \to + \infty $.

Đặt $x$ làm nhân tử chung:

$\mathop {\lim }\limits_{x \to + \infty } {\mkern 1mu} \left( {\sqrt {{x^2} + 1} + x} \right) = \mathop {\lim }\limits_{x \to + \infty } {\mkern 1mu} x\left( {\sqrt {1 + \frac{1}{{{x^2}}}} + 1} \right) = + \infty $

vì $\left\{ \begin{gathered}
\mathop {\lim }\limits_{x \to + \infty } {\mkern 1mu} x = + \infty \hfill \\
\mathop {\lim }\limits_{x \to {2^ + }} {\mkern 1mu} \sqrt {1 + \frac{1}{{{x^2}}}} + 1 = 2 > 0 \hfill \\
\end{gathered} \right..$

Chọn B

Giải nhanh: $x \to + \infty :{\mkern 1mu} {\mkern 1mu} \sqrt[3]{{3{x^3} – 1}} + \sqrt {{x^2} + 2} \sim\sqrt[3]{{3{x^3}}} + \sqrt {{x^2}} $

$ = \left( {\sqrt[3]{3} + 1} \right)x \to + \infty $

Đặt $x$ làm nhân tử chung:

$\mathop {\lim }\limits_{x \to + \infty } {\mkern 1mu} \left( {\sqrt[3]{{3{x^3} – 1}} + \sqrt {{x^2} + 2} } \right) = \mathop {\lim }\limits_{x \to + \infty } {\mkern 1mu} x\left( {\sqrt[3]{{3 – \frac{1}{{{x^3}}}}} + \sqrt {1 + \frac{2}{{{x^2}}}} } \right) = + \infty $

vì $\left\{ \begin{gathered}
\mathop {\lim }\limits_{x \to + \infty } {\mkern 1mu} x = + \infty \hfill \\
\mathop {\lim }\limits_{x \to + \infty } {\mkern 1mu} \left( {\sqrt[3]{{3 – \frac{1}{{{x^3}}}}} + \sqrt {1 + \frac{2}{{{x^2}}}} } \right) = \sqrt[3]{3} + 1 > 0 \hfill \\
\end{gathered} \right..$

Chọn D

Đặt ${x^2}$ làm nhân tử chung:

$\mathop {\lim }\limits_{x \to + \infty } {\mkern 1mu} x\left( {\sqrt {4{x^2} + 7x} + 2x} \right) = \mathop {\lim }\limits_{x \to + \infty } {\mkern 1mu} {x^2}\left( {\sqrt {4 + \frac{7}{x}} + 2} \right) = + \infty $

vì $\left\{ \begin{gathered}
\mathop {\lim }\limits_{x \to + \infty } {\mkern 1mu} {x^2} = + \infty \hfill \\
\mathop {\lim }\limits_{x \to + \infty } {\mkern 1mu} \left( {\sqrt {4 + \frac{7}{x}} + 2} \right) = 4 > 0 \hfill \\
\end{gathered} \right..$

Giải nhanh: $x \to + \infty :x\left( {\sqrt {4{x^2} + 7x} + 2x} \right)\simx\left( {\sqrt {4{x^2}} + 2x} \right)$

$ = 4{x^2} \to + \infty $

Chọn C

Ta có $\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} – 8}}{{{x^2} – 4}} = \mathop {\lim }\limits_{x \to 2} \frac{{(x – 2)({x^2} + 2x + 4)}}{{(x – 2)(x + 2)}}$

$ = \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + 2x + 4}}{{x + 2}} = \frac{{12}}{4} = 3$

Chọn D

$\mathop {\lim }\limits_{x \to – 1} \,\,\frac{{{x^5} + 1}}{{{x^3} + 1}} = \mathop {\lim }\limits_{x \to – 1} \frac{{\left( {x + 1} \right)\left( {{x^4} – {x^3} + {x^2} – x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}}$

$ = \mathop {\lim }\limits_{x \to – 1} \frac{{{x^4} – {x^3} + {x^2} – x + 1}}{{{x^2} – x + 1}} = \frac{5}{3}$

Chọn A

Ta có $\mathop {\lim }\limits_{x \to – \sqrt 3 } \frac{{2{x^3} + 3\sqrt 3 }}{{3 – {x^2}}}$

$ = \mathop {\lim }\limits_{x \to – \sqrt 3 } \frac{{2\left( {x + \sqrt 3 } \right)\left( {{x^2} – \sqrt 3 x + 3} \right)}}{{\left( {\sqrt 3 – x} \right)\left( {\sqrt 3 + x} \right)}}$

$ = \mathop {\lim }\limits_{x \to – \sqrt 3 } \frac{{2\left( {{x^2} – \sqrt 3 x + 3} \right)}}{{\sqrt 3 – x}}$

$ = \frac{{2\left[ {{{\left( { – \sqrt 3 } \right)}^2} – \sqrt 3 .\left( { – \sqrt 3 } \right) + 3} \right]}}{{\sqrt 3 – \left( { – \sqrt 3 } \right)}} = \frac{{18}}{{2\sqrt 3 }} = 3\sqrt 3 $

$\xrightarrow[{}]{}\left\{ \begin{gathered}
a = 3 \hfill \\
b = 0 \hfill \\
\end{gathered} \right. \Rightarrow {a^2} + {b^2} = 9$

Chọn C

$\mathop {\lim }\limits_{x \to – 3} \left| {\frac{{ – {x^2} – x + 6}}{{{x^2} + 3x}}} \right| = \mathop {\lim }\limits_{x \to – 3} \left| {\frac{{\left( {x + 3} \right)\left( {x – 2} \right)}}{{x\left( {x + 3} \right)}}} \right|$

$ = \mathop {\lim }\limits_{x \to – 3} \left| {\frac{{x – 2}}{x}} \right| = \left| {\frac{{ – 3 – 2}}{{ – 3}}} \right| = \frac{5}{3}$

Chọn B

Ta có $3 – x > 0$ với mọi $x < 3,$ do đó:

$\mathop {\lim }\limits_{x \to {3^ – }} \frac{{3 – x}}{{\sqrt {27 – {x^3}} }} = \mathop {\lim }\limits_{x \to {3^ – }} \frac{{3 – x}}{{\sqrt {\left( {3 – x} \right)\left( {9 + 3x + {x^2}} \right)} }}$

$ = \mathop {\lim }\limits_{x \to {3^ – }} \frac{{\sqrt {3 – x} }}{{\sqrt {9 + 3x + {x^2}} }} = \frac{{\sqrt {3 – 3} }}{{\sqrt {9 + 3.3 + {3^2}} }} = 0.$

Chọn A

Ta có

$\mathop {\lim }\limits_{x \to 0} \frac{{\left( {{x^2} + {\pi ^{21}}} \right)\sqrt[7]{{1 – 2x}} – {\pi ^{21}}}}{x}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {{x^2} + {\pi ^{21}}} \right)\left( {\sqrt[7]{{1 – 2x}} – 1} \right)}}{x} + \mathop {\lim }\limits_{x \to 0} x = – \frac{{2{\pi ^{21}}}}{7}$

Chọn D

Ta có $\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sqrt {{x^2} + x} – \sqrt x }}{{{x^2}}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\left( {{x^2} + x} \right) – x}}{{{x^2}\left( {\sqrt {{x^2} + x} + \sqrt x } \right)}}$

$ = \mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{{\sqrt {{x^2} + x} + \sqrt x }} = + \infty $

vì $1 > 0$; $\mathop {\lim }\limits_{x \to {0^ + }} \left( {\sqrt {{x^2} + x} + \sqrt x } \right) = 0$ và $\sqrt {{x^2} + x} + \sqrt x > 0$ với mọi $x > 0.$

Chọn C

Ta có $\mathop {\lim }\limits_{x \to 1} \,\,\frac{{\sqrt[3]{x} – 1}}{{\sqrt[3]{{4x + 4}} – 2}}$

$ = \mathop {\lim }\limits_{x \to 1} \frac{{(x – 1)\left( {\sqrt[3]{{{{\left( {4x + 4} \right)}^2}}} + 2\sqrt[3]{{4x + 4}} + 4} \right)}}{{\left( {4x + 4 – 8} \right)\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right)}}$

$ = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt[3]{{{{\left( {4x + 4} \right)}^2}}} + 2\sqrt[3]{{4x + 4}} + 4} \right)}}{{4\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right)}} = \frac{{12}}{{12}} = 1.$

Chọn B

Ta có $\mathop {\lim }\limits_{x \to 0} \frac{{2\sqrt {1 + x} – \sqrt[3]{{8 – x}}}}{x}$

$ = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{2\sqrt {1 + x} – 2}}{x} + \frac{{2 – \sqrt[3]{{8 – x}}}}{x}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \left( {\frac{2}{{\sqrt {x + 1} + 1}} + \frac{1}{{4 + 2\sqrt[3]{{8 – x}} + \sqrt[3]{{{{\left( {8 – x} \right)}^2}}}}}} \right)$

$ = 1 + \frac{1}{{12}} = \frac{{13}}{{12}}$

Chọn A

Ta có $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{ax + 1}} – \sqrt {1 – bx} }}{x} = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sqrt[3]{{ax + 1}} – 1}}{x} + \frac{{1 – \sqrt {1 – bx} }}{x}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{ax}}{{x\left( {\sqrt[3]{{{{\left( {x + 1} \right)}^2}}} + \sqrt[3]{{x + 1}} + 1} \right)}} + \frac{{bx}}{{x\left( {1 + \sqrt {1 – x} } \right)}}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \left( {\frac{a}{{\left( {\sqrt[3]{{{{\left( {x + 1} \right)}^2}}} + \sqrt[3]{{x + 1}} + 1} \right)}} + \frac{b}{{\left( {1 + \sqrt {1 – x} } \right)}}} \right)$

$ = \frac{a}{3} + \frac{b}{2} = 2$

Vậy ta được: $\left\{ \begin{gathered}
a + b = 5 \hfill \\
\frac{a}{3} + \frac{b}{2} = 2 \hfill \\
\end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}
a + b = 5 \hfill \\
2a + 3b = 12 \hfill \\
\end{gathered} \right.$

$ \Leftrightarrow a = 3,\,\,b = 2$

Chọn D

Ta có $\mathop {\lim }\limits_{x \to – \infty } \frac{{2{x^2} + 5x – 3}}{{{x^2} + 6x + 3}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{2 + \frac{5}{x} – \frac{3}{{{x^2}}}}}{{1 + \frac{6}{x} + \frac{3}{{{x^2}}}}} = 2$.

Giải nhanh : khi $x \to – \infty $ thì : $\frac{{2{x^2} + 5x – 3}}{{{x^2} + 6x + 3}} \sim \frac{{2{x^2}}}{{{x^2}}} = 2.$

Chọn C

Ta có: $\mathop {\lim }\limits_{x \to – \infty } \frac{{2{x^3} + 5{x^2} – 3}}{{{x^2} + 6x + 3}} = \mathop {\lim }\limits_{x \to – \infty } x.\frac{{2 + \frac{5}{x} – \frac{3}{{{x^3}}}}}{{1 + \frac{6}{x} + \frac{3}{{{x^2}}}}} = – \infty .$

Giải nhanh : khi $x \to – \infty $ thì : $\frac{{2{x^3} + 5{x^2} – 3}}{{{x^2} + 6x + 3}} \sim \frac{{2{x^3}}}{{{x^2}}} = 2x \to – \infty .$

Chọn C

Ta có: $\mathop {\lim }\limits_{x \to – \infty } \frac{{2{x^3} – 7{x^2} + 11}}{{3{x^6} + 2{x^5} – 5}}$

$ = \mathop {\lim }\limits_{x \to – \infty } \frac{{\frac{2}{{{x^3}}} – \frac{7}{{{x^4}}} + \frac{{11}}{{{x^6}}}}}{{3 + \frac{2}{x} – \frac{5}{{{x^6}}}}} = \frac{0}{3} = 0$

Giải nhanh : khi $x \to – \infty $ thì : $\frac{{2{x^3} – 7{x^2} + 11}}{{3{x^6} + 2{x^5} – 5}} \sim \frac{{2{x^3}}}{{3{x^6}}} = \frac{2}{3}.\frac{1}{{{x^3}}} \to 0.$

Chọn D

Khi $x \to – \infty $ thì $\sqrt {{x^2}} = – x\xrightarrow[{}]{}\sqrt {{x^2} + 1} – x \sim \sqrt {{x^2}} – x = – x – x = – 2x\not = 0$

$\xrightarrow[{}]{}$chia cả tử và mẫu cho $x$, ta được $\mathop {\lim }\limits_{x \to – \infty } \frac{{2x – 3}}{{\sqrt {{x^2} + 1} – x}} = \mathop {\lim }\limits_{x \to – \infty } \frac{{2 – \frac{3}{x}}}{{ – \sqrt {1 + \frac{1}{{{x^2}}}} – 1}} = – 1$.

Chọn B

Khi $x \to + \infty $ thì $\sqrt {{x^2}} = x\xrightarrow[{}]{}\sqrt {{x^2} + 1} – x \sim \sqrt {{x^2}} – x = x – x = 0$

$\xrightarrow[{}]{}$ Nhân lượng liên hợp:

Ta có $\mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {2 – a} \right)x – 3}}{{\sqrt {{x^2} + 1} – x}}$

$ = \mathop {\lim }\limits_{x \to + \infty } \left( {\left( {2 – a} \right)x – 3} \right)\left( {\sqrt {{x^2} + 1} + x} \right)$

$ = \mathop {\lim }\limits_{x \to + \infty } {x^2}\left( {2 – a – \frac{3}{x}} \right)\left( {\sqrt {1 + \frac{1}{{{x^2}}}} + 1} \right)$

Vì $\left\{ \begin{gathered}
\mathop {\lim }\limits_{x \to + \infty } {x^2} = + \infty \hfill \\
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {1 + \frac{1}{{{x^2}}}} + 1} \right) = 4 > 0 \hfill \\
\end{gathered} \right.$

$ \Rightarrow \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {2 – a} \right)x – 3}}{{\sqrt {{x^2} + 1} – x}} = + \infty $

$ \Leftrightarrow \mathop {\lim }\limits_{x \to + \infty } \left( {2 – a – \frac{3}{x}} \right) = 2 – a > 0 \Rightarrow a < 2$.

Giải nhanh : ta có $x \to + \infty \xrightarrow[{}]{}\frac{{2x – 3}}{{\sqrt {{x^2} + 1} – x}}$

$ = \left( {\left( {2 – a} \right)x – 3} \right)\left( {\sqrt {{x^2} + 1} + x} \right) \sim \left( {2 – a} \right)x.\left( {\sqrt {{x^2}} + x} \right)$

$ = 2\left( {2 – a} \right)x \to + \infty \Leftrightarrow a < 2$

Khi đó $P = {a^2} – 2a + 4 = {\left( {a – 1} \right)^2} + 3 \geqslant 3\,$

$P = 3 \Leftrightarrow a = 1 < 2 \Rightarrow {P_{\min }} = 3$

Chọn C

Giải nhanh: khi $x \to – \infty \xrightarrow[{}]{}\frac{{\sqrt {4{x^2} – x + 1} }}{{x + 1}} \sim \frac{{\sqrt {4{x^2}} }}{x} = \frac{{ – 2x}}{x} = – 2.$

Cụ thể: $\mathop {\lim }\limits_{x \to – \infty } \frac{{\sqrt {4{x^2} – x + 1} }}{{x + 1}}$

$ = \mathop {\lim }\limits_{x \to – \infty } \frac{{ – \sqrt {4 – \frac{1}{x} + \frac{1}{{{x^2}}}} }}{{1 + \frac{1}{x}}} = \frac{{ – \sqrt 4 }}{1} = – 2$

Chọn D

Giải nhanh : khi

$x \to + \infty \xrightarrow[{}]{}\frac{{\sqrt {4{x^2} – 2x + 1} + 2 – x}}{{\sqrt {9{x^2} – 3x} + 2x}}$

$ \sim \frac{{\sqrt {4{x^2}} – x}}{{\sqrt {9{x^2}} + 2x}} = \frac{{2x – x}}{{3x + 2x}} = \frac{1}{5}$

Cụ thể : $\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {4{x^2} – 2x + 1} + 2 – x}}{{\sqrt {9{x^2} – 3x} + 2x}}$

$ = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {4 – \frac{2}{x} + \frac{1}{{{x^2}}}} + \frac{2}{x} – 1}}{{\sqrt {9 – \frac{3}{x}} + 2}} = \frac{1}{5}$

Chọn B

Ta phải có $a{x^2} – 3x > 0$ trên $ + \infty .$

Ta có $\mathop {\lim }\limits_{x \to – \infty } x = – \infty ,\,$

$\mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt[3]{{3 – \frac{1}{{{x^3}}}}} – \sqrt {1 + \frac{2}{{{x^2}}}} } \right) = \sqrt[3]{3} – 1 > 0$

Như vậy xem như “tử” là một đa thức bậc 1. Khi đó $ – \infty $ khi và chỉ khi $\sqrt[3]{3} + 1.$ là đa thức bậc 1.

Ta có $\sqrt {a{x^2} – 3x} + bx \sim \sqrt {a{x^2}} + bx$

$ = \left( { – \sqrt a + b} \right)x\xrightarrow[{}]{} – \sqrt a + b\not = 0$

Khi đó $\frac{{\sqrt {4{x^2} – 2x + 1} + 2 – x}}{{\sqrt {a{x^2} – 3x} + bx}} \sim \frac{{ – 3x}}{{( – \sqrt a + b)x}} = \frac{3}{{b – a}} > L$

$ \Leftrightarrow b – \sqrt a > 0 \Rightarrow b > \sqrt a $

Chọn C

Giải nhanh: $x \to – \infty \to \,\frac{{\sqrt[3]{{{x^3} + 2{x^2} + 1}}}}{{\sqrt {2{x^2} + 1} }} \sim \,\frac{{\sqrt[3]{{{x^3}}}}}{{\sqrt {2{x^2}} }}$

$ = \frac{x}{{ – \sqrt 2 x}} = – \frac{1}{{\sqrt 2 }}$

Cụ thể: $\,\mathop {\lim }\limits_{x \to – \infty } \frac{{\sqrt[3]{{{x^3} + 2{x^2} + 1}}}}{{\sqrt {2{x^2} + 1} }} = \,\mathop {\lim }\limits_{x \to – \infty } \frac{{\sqrt[3]{{1 + \frac{2}{x} + \frac{1}{{{x^3}}}}}}}{{ – \sqrt {2 + \frac{1}{{{x^2}}}} }} = – \frac{1}{{\sqrt 2 }}.$

Chọn B

Giải nhanh: $x \to – \infty \to \sqrt {2{x^2} + 1} + ax \sim \sqrt {2{x^2}} + x$

$ = – \sqrt 2 x + ax = \left( {a – \sqrt 2 } \right)x \to + \infty \Leftrightarrow a – \sqrt 2 < 0 \Leftrightarrow a < \sqrt 2 .$

Cụ thể:

vì $\mathop {\lim }\limits_{x \to – \infty } x = – \infty $ nên $\mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {2{x^2} + 1} + ax} \right) = \mathop {\lim }\limits_{x \to – \infty } x\left( { – \sqrt {2 + \frac{1}{{{x^2}}}} + a} \right) = + \infty $

$ \Leftrightarrow \mathop {\lim }\limits_{x \to – \infty } \left( { – \sqrt {2 + \frac{1}{{{x^2}}}} + a} \right) = a – \sqrt 2 < 0 \Leftrightarrow a < \sqrt 2 .$

Chọn D

Giải nhanh : $x \to – \infty \xrightarrow[{}]{}2{x^3} – {x^2} \sim 2{x^3} \to – \infty .$

Cụ thể: $\mathop {\lim }\limits_{x \to – \infty } \left( {2{x^3} – {x^2}} \right) = \mathop {\lim }\limits_{x \to – \infty } {x^3}\left( {2 – \frac{1}{x}} \right) = – \infty $ vì $\left\{ \begin{gathered}
\mathop {\lim }\limits_{x \to – \infty } {x^3} = – \infty \hfill \\
\mathop {\lim }\limits_{x \to – \infty } \left( {2 – \frac{1}{x}} \right) = 2 > 0 \hfill \\
\end{gathered} \right.$

Chọn A

Ta có $\mathop {\lim }\limits_{x \to {2^ – }} \left( {\frac{1}{{x – 2}} – \frac{1}{{{x^2} – 4}}} \right) = \mathop {\lim }\limits_{x \to {2^ – }} \left( {\frac{{x + 2 – 1}}{{{x^2} – 4}}} \right)$

$ = \mathop {\lim }\limits_{x \to {2^ – }} \left( {\frac{{x + 1}}{{{x^2} – 4}}} \right) = – \infty $

Vì $\mathop {\lim }\limits_{x \to {2^ – }} \left( {x + 1} \right) = 3 > 0;\,\,\mathop {\lim }\limits_{x \to {2^ – }} \left( {{x^2} – 4} \right) = 0$ và ${x^2} – 4 < 0$ với mọi $x \in \left( { – 2;2} \right).$

Chọn A

Vì $\left\{ \begin{gathered}
\mathop {\lim }\limits_{x \to {2^ + }} \left( {x – 15} \right) = – 13 < 0 \hfill \\
\mathop {\lim }\limits_{x \to {2^ + }} \left( {x – 2} \right) = 0 \& x – 2 > 0,\forall x > 2 \hfill \\
\end{gathered} \right.$

$ \to \mathop {\lim }\limits_{x \to {2^ + }} \frac{{x – 15}}{{x – 2}} = – \infty $

Chọn B

$\left\{ \begin{gathered}
\mathop {\lim }\limits_{x \to {2^ + }} \sqrt {x + 2} = 2 > 0 \hfill \\
\mathop {\lim }\limits_{x \to {2^ + }} \sqrt {x – 2} = 0 \& \sqrt {x – 2} > 0,\forall x > 2 \hfill \\
\end{gathered} \right.$

$ \to \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {x + 2} }}{{\sqrt {x – 2} }} = + \infty $

Chọn B

Ta có $\left| {x + 2} \right| = x + 2$ với mọi $x > – 2,$ do đó :

$\mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} \frac{{\left| {3x + 6} \right|}}{{x + 2}} = \mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} \frac{{3\left| {x + 2} \right|}}{{x + 2}}$

$ = \mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} \frac{{3\left( {x + 2} \right)}}{{x + 2}} = \mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} 3 = 3$

Chọn C

Ta có $\mathop {\lim }\limits_{x \to {2^ – }} \frac{{\left| {2 – x} \right|}}{{2{x^2} – 5x + 2}} = \mathop {\lim }\limits_{x \to {2^ – }} \frac{{2 – x}}{{\left( {2 – x} \right)\left( {1 – 2x} \right)}}$

$ = \mathop {\lim }\limits_{x \to {2^ – }} \frac{1}{{1 – 2x}} = – \frac{1}{3}$

Chọn C

Ta có $x + 3 > 0$ với mọi $x > – 3,$ nên:

$\,\mathop {\lim }\limits_{x \to – {3^ + }} \frac{{{x^2} + 13x + 30}}{{\sqrt {\left( {x + 3} \right)\left( {{x^2} + 5} \right)} }} = \,\mathop {\lim }\limits_{x \to – {3^ + }} \frac{{\left( {x + 3} \right)\left( {x + 10} \right)}}{{\sqrt {\left( {x + 3} \right)\left( {{x^2} + 5} \right)} }}$

$ = \,\mathop {\lim }\limits_{x \to – {3^ + }} \frac{{\sqrt {x + 3} .\left( {x + 10} \right)}}{{\sqrt {{x^2} + 5} }} = \frac{{\sqrt { – 3 + 3} \left( { – 3 + 7} \right)}}{{\sqrt {{{\left( { – 3} \right)}^2} + 5} }} = 0$

Chọn B

$\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \sqrt {3{x^2} + 1} = \sqrt {{{3.1}^2} + 1} = 2$

Chọn A

$\mathop {\lim }\limits_{x \to {1^ – }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ – }} \frac{{{x^2} + 1}}{{1 – x}} = + \infty $ vì $\left\{ \begin{gathered}
\mathop {\lim }\limits_{x \to {1^ – }} \left( {{x^2} + 1} \right) = 2 \hfill \\
\mathop {\lim }\limits_{x \to {1^ – }} \left( {1 – x} \right) = 0\,\,\& \,\,1 – x > 0\,\,\left( {\forall x < 1} \right) \hfill \\
\end{gathered} \right..$

Chọn C

Ta có $\left\{ \begin{gathered}
\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {{x^2} – 3} \right) = 1 \hfill \\
\mathop {\lim }\limits_{x \to {2^ – }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ – }} \left( {x – 1} \right) = 1 \hfill \\
\end{gathered} \right.$

$ \Rightarrow \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ – }} f\left( x \right) = 1 \Rightarrow \mathop {\lim }\limits_{x \to 2} f\left( x \right) = 1$

Chọn B

Ta có $\left\{ \begin{gathered}
\mathop {\lim }\limits_{x \to {2^ – }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ – }} \left( {ax – 1} \right) = 2a – 1 \hfill \\
\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {\sqrt {x – 2} + 3} \right) = 3 \hfill \\
\end{gathered} \right..$

Khi đó $\mathop {\lim }\limits_{x \to 2} f\left( x \right)$ tồn tại $ \Leftrightarrow \mathop {\lim }\limits_{x \to {2^ – }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)$

$ \Leftrightarrow 2a – 1 = 3 \Leftrightarrow a = 2$

Chọn C

Ta có $\left\{ \begin{gathered}
\mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} \left( {{x^2} – 2x + 3} \right) = 6 \hfill \\
\mathop {\lim }\limits_{x \to {3^ – }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ – }} \left( {3 – 2{x^2}} \right) = – 15 \hfill \\
\end{gathered} \right.$

$\xrightarrow[{}]{}\mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {3^ – }} f\left( x \right)$

$\xrightarrow{{}}$ không tồn tại giới hạn khi $x \to 3.$

Vậy chỉ có khẳng định C sai.

Chọn C

Ta có $\mathop {\lim }\limits_{x \to 1} \,\,\,\left( {\frac{a}{{1 – x}} – \frac{b}{{1 – {x^3}}}} \right)$

$ = \mathop {\lim }\limits_{x \to 1} \frac{{a + ax + a{x^2} – b}}{{1 – {x^3}}} = \mathop {\lim }\limits_{x \to 1} \frac{{a + ax + a{x^2} – b}}{{\left( {1 – x} \right)\left( {1 + x + {x^2}} \right)}}$

Khi đó $\mathop {\lim }\limits_{x \to 1} \,\,\,\left( {\frac{a}{{1 – x}} – \frac{b}{{1 – {x^3}}}} \right)$ hữu hạn $ \Leftrightarrow 1 + a.1 + a{.1^2} – b = 0 \Leftrightarrow 2a – b = – 1.$

Vậy ta có $\left\{ \begin{gathered}
a + b = 4 \hfill \\
2a – b = – 1 \hfill \\
\end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}
a = 1 \hfill \\
b = 3 \hfill \\
\end{gathered} \right.$

$ \Rightarrow L = – \mathop {\lim }\limits_{x \to 1} \,\,\,\left( {\frac{a}{{1 – x}} – \frac{b}{{1 – {x^3}}}} \right)$

$ = – \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} + x – 2}}{{\left( {1 – x} \right)\left( {1 + x + {x^2}} \right)}} = – \mathop {\lim }\limits_{x \to 1} \frac{{ – \left( {x + 2} \right)}}{{1 + x + {x^2}}} = 1$.

Chọn B

Ta có $\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {1 + 2{x^2}} – x} \right) = \mathop {\lim }\limits_{x \to + \infty } x\left( {\sqrt {\frac{1}{{{x^2}}} + 2} – 1} \right) = + \infty $

Vì $\mathop {\lim }\limits_{x \to + \infty } x = + \infty ;\,\,\,\,\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {\frac{1}{{{x^2}}} + 2} – 1} \right) = \sqrt 2 – 1 > 0.$

Giải nhanh : $x \to + \infty \xrightarrow[{}]{}\sqrt {1 + 2{x^2}} – x \sim \sqrt {2{x^2}} – x$

$ = \sqrt 2 x – x = \left( {\sqrt 2 – 1} \right)x \to + \infty $

Chọn A

$x \to + \infty \to \sqrt {{x^2} + 1} – x \sim \sqrt {{x^2}} – x$

$ = x – x = 0\xrightarrow[{}]{}$Nhân lượng liên hợp.

Giải nhanh: $x \to + \infty \xrightarrow[{}]{}\sqrt {{x^2} + 1} – x$

$ = \frac{1}{{\sqrt {{x^2} + 1} + x}} \sim \frac{1}{{\sqrt {{x^2}} + x}} = \frac{1}{{2x}} \to 0$

Cụ thể: $\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 1} – x} \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{1}{{\sqrt {{x^2} + 1} + x}}$

$ = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{1}{x}}}{{\sqrt {1 + \frac{1}{{{x^2}}}} + 1}} = \frac{0}{2} = 0$

Chọn A

$x \to – \infty \xrightarrow[{}]{}\sqrt {5{x^2} + 2x} + x\sqrt 5 $

$ \sim \sqrt {5{x^2}} + x\sqrt 5 = – \sqrt 5 x + x\sqrt 5 = 0$

$\xrightarrow[{}]{}$Nhân lượng liên hợp:

Giải nhanh: $x \to – \infty \xrightarrow[{}]{}\sqrt {5{x^2} + 2x} + x\sqrt 5 $

$ = \frac{{2x}}{{\sqrt {5{x^2} + 2x} + x\sqrt 5 }} \sim \frac{{2x}}{{\sqrt {5{x^2}} – x\sqrt 5 }}$

$ = \frac{{2x}}{{ – 2\sqrt 5 x}} = – \frac{1}{{\sqrt 5 }}.$

Cụ thể: Ta có $\mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {5{x^2} + 2x} + x\sqrt 5 } \right) = \mathop {\lim }\limits_{x \to – \infty } \frac{{2x}}{{\sqrt {5{x^2} + 2x} + x\sqrt 5 }}$

$ = \mathop {\lim }\limits_{x \to – \infty } \frac{2}{{ – \sqrt {5 + \frac{2}{x}} + \sqrt 5 }} = \frac{2}{{ – 2\sqrt 5 }}$

$ = – \frac{1}{{\sqrt 5 }} = – \frac{1}{5}\sqrt 5 \xrightarrow{{}}\left\{ \begin{gathered}
a = – \frac{1}{5} \hfill \\
b = 0 \hfill \\
\end{gathered} \right. \Rightarrow S = – 1$

Chọn B

. Khi $x \to + \infty \xrightarrow[{}]{}\sqrt {{x^2} + 3x} – \sqrt {{x^2} + 4x} \sim \sqrt {{x^2}} – \sqrt {{x^2}} = 0$

$\xrightarrow[{}]{}$Nhân lượng liên hợp:

Giải nhanh: $x \to + \infty \xrightarrow[{}]{}\sqrt {{x^2} + 3x} – \sqrt {{x^2} + 4x} $

$ = \frac{{ – x}}{{\sqrt {{x^2} + 3x} + \sqrt {{x^2} + 4x} }} \sim \frac{{ – x}}{{\sqrt {{x^2}} + \sqrt {{x^2}} }}$

$ = \frac{{ – x}}{{2x}} = – \frac{1}{2}$

Cụ thể: $\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 3x} – \sqrt {{x^2} + 4x} } \right) = $

$\mathop {\lim }\limits_{x \to + \infty } \frac{{ – x}}{{\sqrt {{x^2} + 3x} + \sqrt {{x^2} + 4x} }}$

$ = \mathop {\lim }\limits_{x \to + \infty } \frac{{ – 1}}{{\sqrt {1 + \frac{3}{x}} + \sqrt {1 + \frac{4}{x}} }} = – \frac{1}{2}$

Chọn D

$\mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt[3]{{3{x^3} – 1}} + \sqrt {{x^2} + 2} } \right)$

$ = \mathop {\lim }\limits_{x \to – \infty } x\left( {\sqrt[3]{{3 – \frac{1}{{{x^3}}}}} – \sqrt {1 + \frac{2}{{{x^2}}}} } \right) = – \infty $

Vì $\mathop {\lim }\limits_{x \to – \infty } x = – \infty ,\,\,\,\,\,\mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt[3]{{3 – \frac{1}{{{x^3}}}}} – \sqrt {1 + \frac{2}{{{x^2}}}} } \right) = \sqrt[3]{3} – 1 > 0.$

Giải nhanh:

$x \to – \infty \to \sqrt[3]{{3{x^3} – 1}} + \sqrt {{x^2} + 2} \sim \sqrt[3]{{3{x^3}}} + \sqrt {{x^2}} $

$ = \left( {\sqrt[3]{3} – 1} \right)x \to – \infty $

Chọn A

Khi $x \to + \infty \xrightarrow[{}]{}\sqrt {{x^2} + x} – \sqrt[3]{{{x^3} – {x^2}}} \sim \sqrt {{x^2}} – – \sqrt[3]{{{x^3}}} = x – x = 0$

$\xrightarrow[{}]{}$Nhân lượng liên hợp:

$\mathop {\lim }\limits_{x \to + \infty } \,\,\left( {\sqrt {{x^2} + x} – \sqrt[3]{{{x^3} – {x^2}}}} \right) = \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + x} – x + x – \sqrt[3]{{{x^3} – {x^2}}}} \right)$

$ = \mathop {\lim }\limits_{x \to + \infty } \left( {\frac{x}{{\sqrt {{x^2} + 1} + x}} + \frac{{{x^2}}}{{{x^2} + x\sqrt[3]{{{x^3} – 1}} + \sqrt[3]{{{{\left( {{x^3} – 1} \right)}^2}}}}}} \right)$

$ = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$

Giải nhanh: $\sqrt {{x^2} + x} – \sqrt[3]{{{x^3} – {x^2}}} = \left( {\sqrt {{x^2} + x} – x} \right) + \left( {x – \sqrt[3]{{{x^3} – {x^2}}}} \right)$

$ = \frac{x}{{\sqrt {{x^2} + 1} + x}} + \frac{{{x^2}}}{{{x^2} + x\sqrt[3]{{{x^3} – 1}} + \sqrt[3]{{{{\left( {{x^3} – 1} \right)}^2}}}}}$

$ \sim \frac{x}{{\sqrt {{x^2}} + x}} + \frac{{{x^2}}}{{{x^2} + x\sqrt[3]{{{x^3}}} + \sqrt[6]{{{x^6}}}}}$

$ = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}\,\,\left( {x \to + \infty } \right).$

Chọn A

$x \to + \infty \xrightarrow[{}]{}\sqrt[3]{{2x – 1}} – \sqrt[3]{{2x + 1}} \sim \sqrt[3]{{2x}} – \sqrt[3]{{2x}} = 0\xrightarrow[{}]{}$nhân lượng liên hợp:

$\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt[3]{{2x – 1}} – \sqrt[3]{{2x + 1}}} \right)$

$ = \mathop {\lim }\limits_{x \to + \infty } \frac{{ – 2}}{{\sqrt[3]{{{{\left( {2x – 1} \right)}^2}}} + \sqrt[3]{{\left( {2x – 1} \right)\left( {2x + 1} \right)}} + \sqrt[3]{{{{\left( {2x + 1} \right)}^2}}}}} = 0$

Giải nhanh: $\sqrt[3]{{2x – 1}} – \sqrt[3]{{2x + 1}} = $

$\frac{{ – 2}}{{\sqrt[{3\sqrt[{}]{{}}}]{{{{\left( {2x – 1} \right)}^2}}} + \sqrt[3]{{4{x^2} – 1}} – \sqrt[3]{{{{\left( {2x + 1} \right)}^2}}}}}$

$ \sim \frac{{ – 2}}{{\sqrt[3]{{4{x^2}}} + \sqrt[3]{{4{x^2}}} + \sqrt[3]{{4{x^2}}}}} = \frac{{ – 2}}{{3\sqrt[3]{{4{x^2}}}}} \to 0$

Chọn B

Ta có $\mathop {\lim }\limits_{x \to 0} \left[ {x\left( {1 – \frac{1}{x}} \right)} \right] = \mathop {\lim }\limits_{x \to 0} \left( {x – 1} \right) = 0 – 1 = – 1.$

Chọn C

Ta có $\,\mathop {\lim }\limits_{x \to {2^ + }} \left( {x – 2} \right)\sqrt {\frac{x}{{{x^2} – 4}}} $

$ = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {x – 2} .\sqrt x }}{{\sqrt {x + 2} }} = \frac{{0.\sqrt 2 }}{2} = 0$

Chọn B

$\mathop {\lim }\limits_{x \to + \infty } x\sqrt {\frac{{2x + 1}}{{3{x^3} + {x^2} + 2}}} = \mathop {\lim }\limits_{x \to + \infty } \sqrt {\frac{{{x^2}\left( {2x + 1} \right)}}{{3{x^3} + {x^2} + 2}}} $

$ = \mathop {\lim }\limits_{x \to + \infty } \sqrt {\frac{{2 + \frac{1}{x}}}{{3 + \frac{1}{x} + \frac{2}{{{x^3}}}}}} = \frac{{\sqrt 6 }}{3}$

Giải nhanh:

$x \to + \infty \xrightarrow[{}]{}x\sqrt {\frac{{2x + 1}}{{3{x^3} + {x^2} + 2}}} $

$ \sim x.\sqrt {\frac{{2x}}{{3{x^2}}}} = \frac{{\sqrt 6 }}{3}.x.\frac{1}{{\sqrt {{x^2}} }} = \frac{{\sqrt 6 }}{3}.x.\frac{1}{x} = \frac{{\sqrt 6 }}{3}$

Chọn B

Ta có $\mathop {\lim }\limits_{x \to 0} {x^2}\left( {\sin \pi x – \frac{1}{{{x^2}}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {{x^2}\sin \pi x – 1} \right) = – 1.$

Chọn C

Với $x \in \left( { – 1;0} \right)$ thì $x + 1 > 0$ và $\frac{x}{{x – 1}} > 0$.

Do đó $\mathop {\lim }\limits_{x \to {{\left( { – 1} \right)}^ + }} \left( {{x^3} + 1} \right)\sqrt {\frac{x}{{{x^2} – 1}}} $

$ = \mathop {\lim }\limits_{x \to {{\left( { – 1} \right)}^ + }} \left( {x + 1} \right)\left( {{x^2} – x + 1} \right)\sqrt {\frac{x}{{\left( {x – 1} \right)\left( {x + 1} \right)}}} $

$ = \mathop {\lim }\limits_{x \to {{\left( { – 1} \right)}^ + }} \sqrt {x + 1} \left( {{x^2} – x + 1} \right)\sqrt {\frac{x}{{x – 1}}} = 0$