Giới hạn của hàm số

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {2^+}} 2x - 5 =  - 1 < 0\\\mathop {\lim }\limits_{x \to {2^+}} \left( {x - 2} \right) = 0;\\x - 2 > 0\forall x > 2\end{array}\)

\( \Rightarrow \mathop {\lim }\limits_{x \to {2^+}} \dfrac{{2x - 5}}{{x - 2}} =  - \infty \)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} [f(x) - g(x)]\\ = \mathop {\lim }\limits_{x \to 1} f(x) - \mathop {\lim }\limits_{x \to 1} g(x)\\ =  - 5 - 2 =  - 7\end{array}\)

\(\mathop {\lim }\limits_{x \to  - \infty } \left( { - {x^3} + 3x} \right) \)\(= \mathop {\lim }\limits_{x \to  - \infty } {x^3}.\left( { - 1 + \dfrac{3}{{{x^2}}}} \right) \)

Vì \(\mathop {\lim }\limits_{x \to  - \infty } {x^3} =  - \infty \)\(\mathop {\lim }\limits_{x \to  - \infty } \left( { - 1 + \dfrac{3}{{{x^2}}}} \right) =  - 1 < 0\)

\(\Rightarrow\mathop {\lim }\limits_{x \to  - \infty } {x^3}.\left( { - 1 + \dfrac{3}{{{x^2}}}} \right)=  + \infty \)

\\(\Rightarrow\mathop {\lim }\limits_{x \to  - \infty } \left( { - {x^3} + 3x} \right)= + \infty \)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {3^ + }} 2x - 1 = 5 > 0\\\mathop {\lim }\limits_{x \to {3^ + }} 3 - x = 0\\3 - x < 0\,khi \,x \to {3^ + }\\ \Rightarrow \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{2x - 1}}{{3 - x}} =  - \infty \end{array}\)

Ta có: \(\mathop {\lim }\limits_{x \to {x_0}} x = {x_0},\mathop {\lim }\limits_{x \to {x_0}} c = c\)  với \(c\) là hằng số.

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

Định lý: Giả sử \(\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = L,\mathop {\lim }\limits_{x \to {x_0}} g\left( x \right) = M\). Khi đó:

+) \(\mathop {\lim }\limits_{x \to {x_0}} \left[ {f\left( x \right) + g\left( x \right)} \right] = L + M\)

+) \(\mathop {\lim }\limits_{x \to {x_0}} \left[ {f\left( x \right) - g\left( x \right)} \right] = L - M\)

+) \(\mathop {\lim }\limits_{x \to {x_0}} \left[ {f\left( x \right).g\left( x \right)} \right] = L.M\)

+) \(\mathop {\lim }\limits_{x \to {x_0}} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{L}{M}\) với \(M \ne 0\)

Từ định lý trên ta thấy đáp án A sai vì thiếu điều kiện \(M \ne 0\).

Ta có \(\mathop {\lim }\limits_{x \to  - 1} \,\,\dfrac{{\left| {x - 1} \right|}}{{{x^4} + x - 3}} = \dfrac{{\left| { - 1 - 1} \right|}}{{1 - 1 - 3}} =  - \dfrac{2}{3}\)

Khi \(x \to x_0^ - \) ta hiểu là \(x\) tiến dần về \({x_0}\) và \(x < {x_0}\).

Ta có x<0 nên:

\(\mathop {\lim }\limits_{x \to  - \infty } \left( {{{\left| x \right|}^3} + 2{x^2} + 3\left| x \right|} \right) \) \(= \mathop {\lim }\limits_{x \to  - \infty } \left( { - {x^3} + 2{x^2} - 3x} \right) \) \(= \mathop {\lim }\limits_{x \to  - \infty } {x^3}\left( { - 1 + \dfrac{2}{x} - \dfrac{3}{{{x^2}}}} \right) =  + \infty .\)

vì \(\mathop {\lim }\limits_{x \to - \infty } {x^3} = - \infty ,\) \(\mathop {\lim }\limits_{x \to - \infty } \left( { - 1 + \frac{2}{x} - \frac{3}{{{x^2}}}} \right) = - 1 < 0\)

Ta có: \(\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = L \Leftrightarrow \mathop {\lim }\limits_{x \to x_0^ + } f\left( x \right) = \mathop {\lim }\limits_{x \to x_0^ - } f\left( x \right) = L\)

Vì \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {2^ + }} \sqrt {x + 2}  = 2 > 0\\\mathop {\lim }\limits_{x \to {2^ + }} \sqrt {x - 2}  = 0\\\sqrt {x - 2}  > 0,\forall x > 2\end{array} \right.\) \( \Rightarrow \mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{\sqrt {x + 2} }}{{\sqrt {x - 2} }} =  + \infty \)

Ta có: \(\mathop {\lim }\limits_{x \to  + \infty } {x^k} =  + \infty \) với \(k\) nguyên dương.

Ta có: \(\mathop {\lim }\limits_{x \to  + \infty } {x^k} =  + \infty \) với \(k\) nguyên dương.

\(\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt[3]{{3{x^3} - 1}} + \sqrt {{x^2} + 2} } \right) = \mathop {\lim }\limits_{x \to  + \infty } x\left( {\sqrt[3]{{3 - \dfrac{1}{{{x^3}}}}} + \sqrt {1 + \dfrac{2}{{{x^2}}}} } \right) =  + \infty \) vì \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to  + \infty } x =  + \infty \\\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt[3]{{3 - \dfrac{1}{{{x^3}}}}} + \sqrt {1 + \dfrac{2}{{{x^2}}}} } \right) = \sqrt[3]{3} + 1 > 0\end{array} \right..\)

Ta có: \(\mathop {\lim }\limits_{x \to  - \infty } {x^4} =  + \infty \) nên A đúng, B sai.

\(\mathop {\lim }\limits_{x \to  + \infty } \left( { - {x^4}} \right) =  - \infty \) nên C và D đều sai.

$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{{x^2} + 1}}{{1 - x}} =  + \infty $ vì \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {1^ - }} \left( {{x^2} + 1} \right) = 2\\\mathop {\lim }\limits_{x \to {1^ - }} \left( {1 - x} \right) = 0\\1 - x > 0\,\,\left( {\forall x < 1} \right)\end{array} \right.\)

\( \Rightarrow \mathop {\lim }\limits_{x \to {x_0}} [3f(x) - 4g(x)] = 3.3 - 4.4 =  - 7\)

\(\mathop {\lim }\limits_{x \to 1} \left( {2x + 1} \right) = 2.1 + 1 = 3\)

Bước 1:

\(\mathop {\lim }\limits_{x \to  - 1} \dfrac{{{x^2} + 2x + 1}}{{2{x^3} + 2}}\)\( = \mathop {\lim }\limits_{x \to  - 1} \dfrac{{{{\left( {x + 1} \right)}^2}}}{{2\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\)

Bước 2:

\( = \mathop {\lim }\limits_{x \to  - 1} \dfrac{{x + 1}}{{2\left( {{x^2} - x + 1} \right)}}\)

Bước 3:

\( = \dfrac{{ - 1 + 1}}{{2.\left( {1 - 1 + 1} \right)}} = 0\)