Hướng dẫn giải (chi tiết)

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

Đáp án cần chọn là: D.

Hướng dẫn giải (chi tiết)

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

Đáp án cần chọn là: B.

Hướng dẫn giải (chi tiết)

$f(x)=\sqrt{x^{2}+2 x+4}-\sqrt{x^{2}-2 x+4}.$

Ta có:

$\begin{array}{l}{\mathop {\lim }\limits_{x \to + \infty } f(x)=\lim\limits _{x \rightarrow+\infty}\left(\sqrt{x^{2}+2 x+4}-\sqrt{x^{2}-2 x+4}\right)} \\ {=\mathop {\lim }\limits_{x \to + \infty }\dfrac{\left(\sqrt{x^{2}+2 x+4}-\sqrt{x^{2}-2 x+4}\right)\left(\sqrt{x^{2}+2 x+4}+\sqrt{x^{2}-2 x+4}\right)}{\left(\sqrt{x^{2}+2 x+4}+\sqrt{x^{2}-2 x+4}\right)}} \\ {=\mathop {\lim }\limits_{x \to + \infty } \dfrac{\left(x^{2}+2 x+4\right)-\left(x^{2}-2 x+4\right)}{\sqrt{x^{2}+2 x+4}+\sqrt{x^{2}-2 x+4}}=\mathop {\lim }\limits_{x \to + \infty } \dfrac{4 x}{\sqrt{x^{2}+2 x+4}+\sqrt{x^{2}-2 x+4}}}\end{array}$

$\begin{array}{l}{=\mathop {\lim }\limits_{x \to + \infty } \dfrac{4}{\sqrt{1+\dfrac{2}{x}+\dfrac{4}{x^{2}}}+\sqrt{1-\dfrac{2}{x}+\dfrac{4}{x^{2}}}}=2} \\ {\mathop {\lim }\limits_{x \to - \infty } f(x)=\lim\limits _{x \rightarrow-\infty}\left(\sqrt{x^{2}+2 x+4}-\sqrt{x^{2}-2 x+4}+\sqrt{x^{2}-2 x+4}\right)} \\ {=\mathop {\lim }\limits_{x \to - \infty } \dfrac{\left(\sqrt{x^{2}+2 x+4}-\sqrt{x^{2}-2 x+4}\right)\left(\sqrt{x^{2}+2 x+4}+\sqrt{x^{2}-2 x+4}\right)}{\left(\sqrt{x^{2}+2 x+4}+\sqrt{x^{2}-2 x+4}\right)}}\end{array}$

$\begin{array}{l}{=\mathop {\lim }\limits_{x \to - \infty } \ \dfrac{\left(x^{2}+2 x+4\right)-\left(x^{2}-2 x+4\right)}{\sqrt{x^{2}+2 x+4}+\sqrt{x^{2}-2 x+4}}=\mathop {\lim }\limits_{x \to - \infty }\dfrac{4 x}{\sqrt{x^{2}+2 x+4}+\sqrt{x^{2}-2 x+4}}} \\ {=\mathop {\lim }\limits_{x \to - \infty } \dfrac{\dfrac{4 x}{x}}{\dfrac{x^{2}+2 x+4}{x}+\dfrac{\sqrt{x^{2}-2 x+4}}{x}}} \\ {=\mathop {\lim }\limits_{x \to - \infty } \dfrac{4}{-\sqrt{1+\dfrac{2}{x}+\dfrac{4}{x^{2}}}-\sqrt{1-\dfrac{2}{x}+\dfrac{4}{x^{2}}}}=\dfrac{4}{-1-1}=-2} \\ {\Rightarrow \mathop {\lim }\limits_{x \to + \infty } f(x)=-\mathop {\lim }\limits_{x \to - \infty } f(x)}\end{array}$

Đáp án cần chọn là: D.

Hướng dẫn giải (chi tiết)

Ta có:

$\begin{array}{l}{f(x)=\sqrt{1+2 x} . \sqrt[3]{1+3 x} \cdot \sqrt[4]{1+4 x}-1} \\ {=\sqrt{1+2 x}-\sqrt{1+2 x}+\sqrt{1+2 x} \cdot \sqrt[3]{1+3 x}-\sqrt{1+2 x} \cdot \sqrt[3]{1+3 x}+\sqrt{1+2 x} \cdot \sqrt[3]{1+3 x}}.(\sqrt[4]{{1 + 4x}} - 1) \\ {=(\sqrt{1+2 x}-1)+\sqrt{1+2 x}(\sqrt[3]{1+3 x}-1)+\sqrt{1+2 x} \cdot \sqrt[3]{1+3 x} .(\sqrt[4]{1+4 x}-1)}\end{array}$

$\begin{array}{*{20}{l}} { \Rightarrow \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x} \cdot \sqrt[3]{{1 + 3x}} \cdot \sqrt[4]{{1 + 4x}} - 1}}{x}}\\ { = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sqrt {1 + 2x} - 1}}{x}} \right) + \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} \cdot \dfrac{{\sqrt[3]{{1 + \mathop {\lim }\limits_{x \to 0} 3x}} - 1}}{x}} \right) + \mathop {\lim }\limits_{x \to 0} \;\quad \left( {\sqrt {1 + 2x} \cdot \sqrt[3]{{1 + 3x}} \cdot \dfrac{{\sqrt[4]{{1 + 4x}} - 1}}{x}} \right)} \end{array}$

Tính:

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

$\begin{array}{l} \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} \cdot \dfrac{{\sqrt[3]{{1 + 3x}} - 1}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} .\dfrac{{(\sqrt[3]{{1 + 3x}} - 1)\left[ {{{(\sqrt[3]{{1 + 3x}})}^2} + \sqrt[3]{{1 + 3x}} + 1} \right]}}{{x \cdot \left[ {{{(\sqrt[3]{{1 + 3x}})}^2} + \sqrt[3]{{1 + 3x}} + 1} \right]}}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {1 + 2x} \cdot \dfrac{{3x}}{{x \cdot \left[ {{{(\sqrt[3]{{1 + 3x}})}^2} + \sqrt[3]{{1 + 3x}} + 1} \right]}}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{3\sqrt {1 + 2x} }}{{\left[ {{{(\sqrt[3]{{1 + 3x}})}^2} + \sqrt[3]{{1 + 3x}} + 1} \right]}}} \right) = \dfrac{{3.1}}{{1 + 1 + 1}} = 1 \end{array}$

$\mathop {\lim }\limits_{x \to 0} \left(\sqrt{1+2 x} \cdot \sqrt[3]{1+3 x} . \dfrac{\sqrt[4]{1+4 x}-1}{x}\right)\\=\mathop {\lim }\limits_{x \to 0}\left(\sqrt{1+2 x} \cdot \sqrt[3]{1+3 x} \cdot \dfrac{(\sqrt[4]{1+4 x}-1)\left[(\sqrt[4]{1+4 x})^{3}+(\sqrt[4]{1+4 x})^{2}+\sqrt[4]{1+4 x}+1\right]}{x\left[\left(\sqrt[4]{1+4 x )^{3}+(\sqrt[4]{1+4 x})^{2}+\sqrt[4]{1+4 x}+1 ]}\right.\right.}\right)$

$\begin{array}{l}{=\mathop {\lim }\limits_{x \to 0} \left(\sqrt{1+2 x} \sqrt[3]{1+3 x} \cdot \dfrac{4 x}{x\left[(\sqrt[4]{1+4 x})^{3}+(\sqrt[4]{1+4 x})^{2}+\sqrt[4]{1+4 x}+1\right]}\right)} \\ {=\mathop {\lim }\limits_{x \to 0} \dfrac{4 \sqrt{1+2 x} \cdot \sqrt[3]{1+3 x}}{(\sqrt[4]{1+4 x})^{3}+(\sqrt[4]{1+4 x})^{2}+\sqrt[4]{1+4 x}+1}=\dfrac{4.1 .1}{1+1+1+1}=1}\end{array}$

Suy ra $\mathop {\lim }\limits_{x \to 0} f(x)=1+1+1=3$

Đáp án D

Hướng dẫn giải (chi tiết)

Đặt $x=\dfrac{1}{y}$, khi $x \rightarrow+\infty : y \rightarrow 0$

Ta có $f\left( y \right) = \sqrt[n]{{\left( {\dfrac{1}{y} + 1} \right)\left( {\dfrac{1}{y} + 2} \right) \ldots \left( {\dfrac{1}{y} + n} \right)}} - \dfrac{1}{y}$ và $\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = \mathop {\lim }\limits_{y \to 0} f\left( y \right)$

Suy ra 

$\begin{array}{*{20}{l}}\n{\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt[n]{{(x + 1)(x + 2) \ldots (x + n)}} - x} \right) = \mathop {\lim }\limits_{y \to 0} \left( {\sqrt[n]{{\left( {\dfrac{1}{y} + 1} \right)\left( {\dfrac{1}{y} + 2} \right) \ldots \left( {\dfrac{1}{y} + n} \right)}} - \dfrac{1}{y}} \right)}\\\n{ = \mathop {\lim }\limits_{y \to 0} \dfrac{{\sqrt[n]{{(1 + y)(1 + 2y) \ldots (1 + ny)}} - 1}}{y}}\n\end{array}$

$\sqrt[n]{(1+y)(1+2 y) \ldots(1+n y)}-1$

$=\sqrt[n]{1+y}-\sqrt[n]{1+y}+\sqrt[n]{(1+y)(1+2 y)}-\sqrt[n]{(1+y)(1+2 y)}+\ldots$

$+\sqrt[n]{(1+y)(1+2 y) \ldots(1+(n-1) y)}$

$-\sqrt[n]{(1+y)(1+2 y) \ldots(1+(n-1) y)}+\sqrt[n]{(1+y)(1+2 y) \ldots(1+n y)}-1$

$=(\sqrt[n]{1+y}-1)+\sqrt[n]{1+y}(\sqrt[n]{1+2 y}-1)+\ldots$

$+\sqrt[n]{(1+y)(1+2 y) \ldots(1+(n-1) y)}(\sqrt[n]{1+n y}-1)$

$\Rightarrow\mathop {\lim }\limits_{y \to 0} \dfrac{\sqrt[5]{(1+y)(1+2 y) \cdots(1+n y)}-1}{y}\\=\mathop {\lim }\limits_{y \to 0} \left[\dfrac{(\sqrt[n]{1+y}-1)}{y}\right]+\mathop {\lim }\limits_{y \to 0} \left[\sqrt[n]{1+y} \cdot \dfrac{(\sqrt[n]{1+2 y}-1)}{y}\right]+\ldots+\mathop {\lim }\limits_{y \to 0} \left[\sqrt[n]{(1+y)(1+2 y) \ldots(1+(n-1) y)} \cdot \dfrac{(\sqrt[n]{1+n y}-1)}{y}\right]$

Tổng quát:

$\mathop {\lim }\limits_{y \to 0} \left[\sqrt[n]{(1+y)(1+2 y) \ldots(1+(k-1) y)} \cdot \dfrac{\sqrt[n]{1+k y}-1}{y}\right]$

$=\mathop {\lim }\limits_{y \to 0} {\rm{ }}\left( {\sqrt[n]{{(1 + y)(1 + 2y) \ldots (1 + (k - 1)y)}}.\dfrac{{(\sqrt[n]{{1 + ky}} - 1)\left[ {{{(\sqrt[n]{{1 + ky}})}^{n - 1}} + {{(\sqrt[n]{{1 + ky}})}^{n - 2}} + \ldots + 1} \right]}}{{y\left[ {{{(\sqrt[n]{{1 + ky}})}^{n - 1}} + {{(\sqrt[n]{{1 + ky}})}^{n - 2}} + \ldots + 1} \right]}}} \right)$

$=\mathop {\lim }\limits_{y \to 0} \dfrac{(1+k y-1) \cdot \sqrt[n]{(1+y)(1+2 y) \ldots(1+(k-1) y)}}{y(\sqrt[n]{1+k y})^{n-1}+(\sqrt[n]{1+k y})^{n-2}+\ldots+1}$

$=\mathop {\lim }\limits_{y \to 0} \dfrac{k \cdot \sqrt[n]{(1+y)(1+2 y) \ldots(1+(k-1) y)}}{(\sqrt[n]{1+k y})^{n-1}+(\sqrt[n]{1+k y})^{n-2}+\ldots+1}=\dfrac{k}{n}$

Khi đó:

$\begin{array}{l}{\mathop {\lim }\limits_{y \to 0} \dfrac{\sqrt[n]{(1+y)(1+2 y) \ldots(1+n y)}-1}{y}=\dfrac{1}{n}+\dfrac{2}{n}+\dfrac{3}{n}+\ldots+\dfrac{n}{n}=\dfrac{1+2+3+\ldots+n}{n}} \\ {=\dfrac{\dfrac{n(n+1)}{2}}{n}=\dfrac{n+1}{2}}\end{array}$

Đáp án cần chọn là: B.

Hướng dẫn giải (chi tiết)

Đặt $y = \sin x$ khi đó từ $x \to \dfrac{\pi }{2}$, thì  $y \to 1$ và ta viết lại giới hạn L dưới dạng sau:

$\begin{array}{l}\nL = \mathop {\lim }\limits_{x \to 1} \dfrac{{1 - {y^{m + n + p}}}}{{\sqrt[3]{{(1 - {y^m})(1 - {y^n})(1 - {y^p})}}}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{(1 - y)(1 + y + {y^2} + ... + {y^{m + n + p + 1}})}}{{(1 - y)\sqrt[3]{{(1 + y + ... + {y^{m - 1}})(1 + y + ... + {y^{n - 1}})(1 + y + ... + {y^{p - 1}})}}}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{1 + y + {y^2} + ... + {y^{m + n + p + 1}}}}{{\sqrt[3]{{(1 + y + ... + {y^{m - 1}})(1 + y + ... + {y^{n - 1}})(1 + y + ... + {y^{p - 1}})}}}}\\ = \dfrac{{m + n + p}}{{\sqrt[3]{{mnp}}}}\n\end{array}$

Hướng dẫn giải (chi tiết)

Ta có: 

$\begin{array}{l}\nL = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt[{2m}]{{{{\cos }^2}\alpha x}} - \sqrt[{2n}]{{{{\cos }^2}\beta x}}}}{{{{\sin }^2}\gamma x}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{(\sqrt[{2m}]{{{{\cos }^2}\alpha x}} - 1) - (\sqrt[{2n}]{{{{\cos }^2}\beta x}} - 1)}}{{{{\sin }^2}\gamma x}}\\ = \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\sqrt[{2m}]{{{{\cos }^2}\alpha x}} - 1}}{{{{\sin }^2}\alpha x}}.\dfrac{{{{\sin }^2}\alpha x}}{{{{\sin }^2}\gamma x}} - \dfrac{{\sqrt[{2n}]{{{{\cos }^2}\beta x}} - 1}}{{{{\sin }^2}\beta x}}.\dfrac{{{{\sin }^2}\beta x}}{{{{\sin }^2}\gamma x}}} \right].(1)\n\end{array}$

Bằng phép biến đổi, ta có:

$\begin{array}{l}\n\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt[{2m}]{{{{\cos }^2}\alpha x}} - 1}}{{{{\sin }^2}\alpha x}} = \mathop {\lim }\limits_{y \to 0} \dfrac{{\sqrt[{2m}]{{1 - y}} - 1}}{z} = \dfrac{{ - 1}}{{2m}},\\\n\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt[{2n}]{{{{\cos }^2}\beta x}} - 1}}{{{{\sin }^2}\beta x}} = \mathop {\lim }\limits_{z \to 0} \dfrac{{\sqrt[{2n}]{{1 - z}} - 1}}{z} = \dfrac{{ - 1}}{{2n}}.\n\end{array}$

Mặt khác ta lại có: 

 $\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}\alpha x}}{{{{\sin }^2}\gamma x}} = {\left( {\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \alpha x}}{{\sin \gamma x}}} \right)^2} = \dfrac{{{\alpha ^2}}}{{{\gamma ^2}}}$

và tương tự: $\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}\beta x}}{{{{\sin }^2}\gamma x}} = \dfrac{{{\beta ^2}}}{{{\gamma ^2}}}$.

Thay vào (1) , ta được : 

$L = \dfrac{1}{{2{\gamma ^2}}}\left( {\dfrac{{{\beta ^2}}}{n} - \dfrac{{{\alpha ^2}}}{m}} \right)$.

Hướng dẫn giải (chi tiết)

Ta có:

$\begin{array}{l}\nL = \mathop {\lim }\limits_{x \to 0} \dfrac{{({e^{\tan 2x}} - 1) - ({e^{\tan x}} - 1)}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan 2x - 1}}}}{x} - \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan x}} - 1}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^{\tan 2x - 1}}}}{{\tan 2x}}.\dfrac{{\tan 2x}}{x}} \right) - \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^{\tan x}} - 1}}{{\tan x}}.\dfrac{{\tan x}}{x}} \right).(1)\n\end{array}$

Để ý rằng: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan 2x}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1$, và 

$\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan 2x - 1}}}}{{\tan 2x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{y - 1}}}}{y} = 1;\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan x - 1}}}}{{\tan x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{z - 1}}}}{z} = 1.$

Từ đó theo (1) , ta có: $L = 2 - 1 = 1,$

Vậy $\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan 2x}} - {e^{\tan x}}}}{x} = 1$

Hướng dẫn giải (chi tiết)

Thực hiện phép nhân liên hợp ta có:

$\begin{array}{l}\nL = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{x + \sqrt {x + \sqrt x } - x}}{{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x }} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\sqrt {x + \sqrt x } }}{{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x }}\\ = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {1 + \dfrac{{\sqrt x }}{x}} }}{{\sqrt {1 + \frac{{\sqrt {x + \sqrt x } }}{x} + 1} }}(1)\n\end{array}$

Lại có:

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

Vì thế từ (1) , ta suy ra: 

$L = \dfrac{{\sqrt {1 + 0} }}{{\sqrt {1 + 0} + 1}} = \dfrac{1}{2}$.