Giới hạn của hàm số

Bước 1:

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

Bước 2:

$=  + \infty$

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

Bước 1:

$\mathop {\lim }\limits_{x \to  - \infty } \left( {{x^3} - 4{x^5} + 2x + 1} \right) = \mathop {\lim }\limits_{x \to  - \infty } {x^5}\left( { - 4 + \dfrac{1}{{{x^2}}} + \dfrac{2}{{{x^4}}} + \dfrac{1}{{{x^5}}}} \right)$.

Bước 2:

Do $\mathop {\lim }\limits_{x \to  - \infty } {x^5} =  - \infty ;$$\mathop {\lim }\limits_{x \to  - \infty } \left( { - 4 + \dfrac{1}{{{x^2}}} + \dfrac{2}{{{x^4}}} + \dfrac{1}{{{x^5}}}} \right) =  - 4 < 0$

$\Rightarrow \mathop {\lim }\limits_{x \to  - \infty } {x^5}\left( { - 4 + \dfrac{1}{{{x^2}}} + \dfrac{2}{{{x^4}}} + \dfrac{1}{{{x^5}}}} \right) =  + \infty$

$\mathop {\lim }\limits_{x \to {1^ - }} \dfrac{5}{{x - 1}} =  - \infty$

Vì $x - 1 < 0$ và $\mathop {\lim }\limits_{x \to {1^ - }} \left( {x - 1} \right) = 0$ khi $x \to {1^ - }$.

Hàm số $y = f\left( x \right)$ có giới hạn là số $L$ khi $x$ dần tới ${x_0}$ kí hiệu là $\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = L$.

$\mathop {\lim }\limits_{x \to 3} \sqrt {\dfrac{{9{x^2} - x}}{{\left( {2x - 1} \right)\left( {{x^4} - 3} \right)}}}$

$= \sqrt {\dfrac{{{{9.3}^2} - 3}}{{\left( {2.3 - 1} \right)\left( {{3^4} - 3} \right)}}}  = \dfrac{1}{{\sqrt 5 }} = \dfrac{{\sqrt 5 }}{5}$

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

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

Số $L$ là: + giới hạn bên phải của hàm số $y = f\left( x \right)$ kí hiệu là $\mathop {\lim }\limits_{x \to x_0^ + } f\left( x \right) = L$

               + giới hạn bên trái của hàm số $y = f\left( x \right)$ kí hiệu là $\mathop {\lim }\limits_{x \to x_0^ - } f\left( x \right) = L$

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

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 $\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {2^ + }} \left( {x - 15} \right) =  - 13 < 0\\\mathop {\lim }\limits_{x \to {2^ + }} \left( {x - 2} \right) = 0\\x - 2 > 0,\forall x > 2\end{array} \right.$ $\Rightarrow \mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{x - 15}}{{x - 2}} =  - \infty$

Ta có: $\mathop {\lim }\limits_{x \to  \pm \infty } c = c,\mathop {\lim }\limits_{x \to  \pm \infty } \dfrac{c}{{{x^k}}} = 0$ nên đáp án A đúng.

Ta có: $\mathop {\lim }\limits_{x \to  + \infty } f\left( x \right) =  + \infty  \Leftrightarrow \mathop {\lim }\limits_{x \to  + \infty } \left[ { - f\left( x \right)} \right] =  - \infty$

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

Ta có: $\mathop {\lim }\limits_{x \to  - \infty } {x^k} =  + \infty$ nếu $k$ chẵn và $\mathop {\lim }\limits_{x \to  - \infty } {x^k} =  - \infty$ nếu $k$ lẻ.

Do đó, vì $n = 2k + 1,k \in N$ là số nguyên dương lẻ nên $\mathop {\lim }\limits_{x \to  - \infty } {x^n} =  - \infty$

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

$\begin{array}{l}\mathop {\lim }\limits_{x \to  - \infty } \left( {{x^2} + 3x - 1} \right)\\ = \mathop {\lim }\limits_{x \to  - \infty } {x^2}\left( {1 + \dfrac{3}{x} - \dfrac{1}{{{x^2}}}} \right) =  + \infty \end{array}$

Bước 1:

$\begin{array}{l}a{x^2} + bx - 5\\ = (ax + a + b)(x - 1) + a + b - 5\end{array}$

Bước 2:

$\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \dfrac{{a{x^2} + bx - 5}}{{x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \left( {ax + a + b + \dfrac{{a + b - 5}}{{x - 1}}} \right) = 20\\ \Leftrightarrow \left\{ \begin{array}{l}a.1 + b + a = 20\\a + b - 5 = 0\end{array} \right.\end{array}$

$\begin{array}{l} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{a = 15}\\{6 =  - 10}\end{array}} \right.\\ \Rightarrow P = {a^2} + {b^2} - a - b = 320\end{array}$

Bước 1:

Đặt $g\left( x \right) = \dfrac{{f\left( x \right) - 2}}{{x - 1}} \Rightarrow f\left( x \right) = \left( {x - 1} \right)g\left( x \right) + 2$

$\Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \left[ {\left( {x - 1} \right)g\left( x \right) + 2} \right] = 2$.

Bước 2:

Ta có:

$\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 2}}{{\left( {{x^2} - 1} \right)\left[ {f\left( x \right) + 1} \right]}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 2}}{{x - 1}}.\dfrac{1}{{\left( {x + 1} \right)\left[ {f\left( x \right) + 1} \right]}}\\ = 12.\dfrac{1}{{2.\left( {2 + 1} \right)}} = 2\end{array}$