GIỚI HẠN LƯỢNG GIÁC
Phương pháp:
Ta sử dụng các công thức lượng giác biến đổi về các dạng sau:
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{x}{{\sin x}} = 1$, từ đây suy ra$\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{x}{{\tan x}} = 1$.
Nếu $\mathop {\lim }\limits_{x \to {x_0}} u(x) = 0 \Rightarrow \mathop {\lim }\limits_{x \to {x_0}} \frac{{\sin u(x)}}{{u(x)}} = 1$ và$\mathop {\lim }\limits_{x \to {x_0}} \frac{{\tan u(x)}}{{u(x)}} = 1$.
Phương pháp:
Ta sử dụng các công thức lượng giác biến đổi về các dạng sau:
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{x}{{\sin x}} = 1$, từ đây suy ra$\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{x}{{\tan x}} = 1$.
Nếu $\mathop {\lim }\limits_{x \to {x_0}} u(x) = 0 \Rightarrow \mathop {\lim }\limits_{x \to {x_0}} \frac{{\sin u(x)}}{{u(x)}} = 1$ và$\mathop {\lim }\limits_{x \to {x_0}} \frac{{\tan u(x)}}{{u(x)}} = 1$.
Ví dụ vận dụng
Câu 1. Tìm giới hạn $A = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos ax}}{{{x^2}}}$ :
A. + ∞
B. - ∞
C. \(\frac{a}{2}\)
D. 0
Chọn C
Ta có:$A = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}\frac{{ax}}{2}}}{{{x^2}}} = \frac{a}{2}\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{\sin \frac{{ax}}{2}}}{{\frac{{ax}}{2}}}} \right)^2} = \frac{a}{2}$.
Ta có:$A = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}\frac{{ax}}{2}}}{{{x^2}}} = \frac{a}{2}\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{\sin \frac{{ax}}{2}}}{{\frac{{ax}}{2}}}} \right)^2} = \frac{a}{2}$.
A. + ∞
B. - ∞
C. \(\frac{m}{n}\)
D. 0
Chọn C
Ta có: $\frac{{1 + \sin mx - \cos mx}}{{1 + \sin nx - \cos nx}} = \frac{{2{{\sin }^2}\frac{{mx}}{2} + 2\sin \frac{{mx}}{2}\cos \frac{{mx}}{2}}}{{2{{\sin }^2}\frac{{nx}}{2} + 2\sin \frac{{nx}}{2}\cos \frac{{nx}}{2}}}$
$ = \frac{m}{n}\frac{{\sin \frac{{mx}}{2}}}{{\frac{{mx}}{2}}}.\frac{{\frac{{nx}}{2}}}{{\sin \frac{{nx}}{2}}}.\frac{{\sin \frac{{mx}}{2} + \cos \frac{{mx}}{2}}}{{\sin \frac{{nx}}{2} + \cos \frac{{nx}}{2}}}$
$A = \frac{m}{n}\mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{{mx}}{2}}}{{\frac{{mx}}{2}}}.\mathop {\lim }\limits_{x \to 0} \frac{{\frac{{nx}}{2}}}{{\sin \frac{{nx}}{2}}}.\mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{{mx}}{2} + \cos \frac{{mx}}{2}}}{{\sin \frac{{nx}}{2} + \cos \frac{{nx}}{2}}} = \frac{m}{n}$.
Ta có: $\frac{{1 + \sin mx - \cos mx}}{{1 + \sin nx - \cos nx}} = \frac{{2{{\sin }^2}\frac{{mx}}{2} + 2\sin \frac{{mx}}{2}\cos \frac{{mx}}{2}}}{{2{{\sin }^2}\frac{{nx}}{2} + 2\sin \frac{{nx}}{2}\cos \frac{{nx}}{2}}}$
$ = \frac{m}{n}\frac{{\sin \frac{{mx}}{2}}}{{\frac{{mx}}{2}}}.\frac{{\frac{{nx}}{2}}}{{\sin \frac{{nx}}{2}}}.\frac{{\sin \frac{{mx}}{2} + \cos \frac{{mx}}{2}}}{{\sin \frac{{nx}}{2} + \cos \frac{{nx}}{2}}}$
$A = \frac{m}{n}\mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{{mx}}{2}}}{{\frac{{mx}}{2}}}.\mathop {\lim }\limits_{x \to 0} \frac{{\frac{{nx}}{2}}}{{\sin \frac{{nx}}{2}}}.\mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{{mx}}{2} + \cos \frac{{mx}}{2}}}{{\sin \frac{{nx}}{2} + \cos \frac{{nx}}{2}}} = \frac{m}{n}$.
A. + ∞
B. - ∞
C. 3
D. 0
Chọn C
Ta có:
$\begin{array}{l} \frac{{1 - \cos x.\cos 2x.\cos 3x}}{{{x^2}}} = \frac{{1 - \cos x + \cos x\cos 2x(1 - \cos 3x) + \cos x(1 - \cos 2x)}}{{{x^2}}}\\ = \frac{{1 - \cos x}}{{{x^2}}} + \cos x.\cos 2x\frac{{1 - \cos 3x}}{{{x^2}}} + \cos x\frac{{1 - \cos 2x}}{{{x^2}}}\\ B = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{{x^2}}} + \mathop {\lim }\limits_{x \to 0} \cos x.\cos 2x\frac{{1 - \cos 3x}}{{{x^2}}} + \mathop {\lim }\limits_{x \to 0} \cos x\frac{{1 - \cos 2x}}{{{x^2}}} = 3 \end{array}$
Ta có:
$\begin{array}{l} \frac{{1 - \cos x.\cos 2x.\cos 3x}}{{{x^2}}} = \frac{{1 - \cos x + \cos x\cos 2x(1 - \cos 3x) + \cos x(1 - \cos 2x)}}{{{x^2}}}\\ = \frac{{1 - \cos x}}{{{x^2}}} + \cos x.\cos 2x\frac{{1 - \cos 3x}}{{{x^2}}} + \cos x\frac{{1 - \cos 2x}}{{{x^2}}}\\ B = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{{x^2}}} + \mathop {\lim }\limits_{x \to 0} \cos x.\cos 2x\frac{{1 - \cos 3x}}{{{x^2}}} + \mathop {\lim }\limits_{x \to 0} \cos x\frac{{1 - \cos 2x}}{{{x^2}}} = 3 \end{array}$
A. + ∞
B. - ∞
C. 1
D. 0
Chọn D
Ta có:$A = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x}}{{\sin \frac{{3x}}{2}}} = \mathop {\lim }\limits_{x \to 0} x{(\frac{{\sin x}}{x})^2}.\frac{3}{2}\mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{{3x}}{2}}}{{\frac{{3x}}{2}}} = 0$.
Ta có:$A = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x}}{{\sin \frac{{3x}}{2}}} = \mathop {\lim }\limits_{x \to 0} x{(\frac{{\sin x}}{x})^2}.\frac{3}{2}\mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{{3x}}{2}}}{{\frac{{3x}}{2}}} = 0$.
A. + ∞
B. - ∞
C. 5/2
D. 0
Chọn C
$B = \mathop {\lim }\limits_{x \to 0} \frac{{2\sin \frac{{5x}}{2}\sin \frac{x}{2}}}{{ - 2x\cos \frac{{7x}}{2}\sin \frac{x}{2}}} = - \mathop {\lim }\limits_{x \to 0} (\frac{5}{2}.\frac{{\sin \frac{{5x}}{2}}}{{\frac{{5x}}{2}}}).\mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos \frac{{7x}}{2}}} = \frac{5}{2}$.
$B = \mathop {\lim }\limits_{x \to 0} \frac{{2\sin \frac{{5x}}{2}\sin \frac{x}{2}}}{{ - 2x\cos \frac{{7x}}{2}\sin \frac{x}{2}}} = - \mathop {\lim }\limits_{x \to 0} (\frac{5}{2}.\frac{{\sin \frac{{5x}}{2}}}{{\frac{{5x}}{2}}}).\mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos \frac{{7x}}{2}}} = \frac{5}{2}$.
A. + ∞
B. - ∞
C. 6
D. 0
Chọn C
$\begin{array}{l} C = \mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^2}2x}}{{1 - \sqrt[3]{{\cos 2x}}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^2}2x(1 + \sqrt[3]{{\cos 2x}} + \sqrt[3]{{{{\cos }^2}2x}})}}{{1 - \cos 2x}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^2}2x(1 + \sqrt[3]{{\cos 2x}} + \sqrt[3]{{{{\cos }^2}2x}})}}{{2{{\sin }^2}x}}\\ = 2\mathop {\lim }\limits_{x \to 0} {(\frac{{\tan 2x}}{{2x}})^2}.{(\frac{x}{{\sin x}})^2}(1 + \sqrt[3]{{\cos 2x}} + \sqrt[3]{{{{\cos }^2}2x}})\\ \Rightarrow C = 6 \end{array}$
$\begin{array}{l} C = \mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^2}2x}}{{1 - \sqrt[3]{{\cos 2x}}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^2}2x(1 + \sqrt[3]{{\cos 2x}} + \sqrt[3]{{{{\cos }^2}2x}})}}{{1 - \cos 2x}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^2}2x(1 + \sqrt[3]{{\cos 2x}} + \sqrt[3]{{{{\cos }^2}2x}})}}{{2{{\sin }^2}x}}\\ = 2\mathop {\lim }\limits_{x \to 0} {(\frac{{\tan 2x}}{{2x}})^2}.{(\frac{x}{{\sin x}})^2}(1 + \sqrt[3]{{\cos 2x}} + \sqrt[3]{{{{\cos }^2}2x}})\\ \Rightarrow C = 6 \end{array}$
A. + ∞
B. - ∞
C. 7/2
D. 0
Chọn C
Ta có: $D = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\frac{{\sqrt {1 + x\sin 3x} - \cos 2x}}{{{x^2}}}}}$
Mà : $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x\sin 3x} - \cos 2x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x\sin 3x} - 1}}{{{x^2}}} + \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos 2x}}{{{x^2}}}$
$ = 3\mathop {\lim }\limits_{x \to 0} (\frac{{\sin 3x}}{{3x}}.\frac{1}{{\sqrt {1 + x\sin 3x} + 1}}) + 2 = \frac{7}{2}$.
Vậy:$D = \frac{7}{2}$.
Ta có: $D = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\frac{{\sqrt {1 + x\sin 3x} - \cos 2x}}{{{x^2}}}}}$
Mà : $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x\sin 3x} - \cos 2x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x\sin 3x} - 1}}{{{x^2}}} + \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos 2x}}{{{x^2}}}$
$ = 3\mathop {\lim }\limits_{x \to 0} (\frac{{\sin 3x}}{{3x}}.\frac{1}{{\sqrt {1 + x\sin 3x} + 1}}) + 2 = \frac{7}{2}$.
Vậy:$D = \frac{7}{2}$.
A. + ∞
B. - ∞
C. \(\frac{n}{m}\)
D. 0
Chọn C
$A = \mathop {\lim }\limits_{x \to 1} \frac{{\sin \pi (1 - {x^m})}}{{\sin \pi (1 - {x^n})}} = \mathop {\lim }\limits_{x \to 1} \frac{{\sin \pi (1 - {x^m})}}{{\pi (1 - {x^m})}}.\mathop {\lim }\limits_{x \to 1} \frac{{\pi (1 - {x^n})}}{{\sin \pi (1 - {x^n})}}.\mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^n}}}{{1 - {x^m}}}$
$ = \mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^n}}}{{1 - {x^m}}} = \mathop {\lim }\limits_{x \to 1} \frac{{(1 - x)({x^{n - 1}} + {x^{n - 2}} + ... + 1)}}{{(1 - x)({x^{m - 1}} + {x^{m - 2}} + ... + 1)}} = \frac{n}{m}.$
$A = \mathop {\lim }\limits_{x \to 1} \frac{{\sin \pi (1 - {x^m})}}{{\sin \pi (1 - {x^n})}} = \mathop {\lim }\limits_{x \to 1} \frac{{\sin \pi (1 - {x^m})}}{{\pi (1 - {x^m})}}.\mathop {\lim }\limits_{x \to 1} \frac{{\pi (1 - {x^n})}}{{\sin \pi (1 - {x^n})}}.\mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^n}}}{{1 - {x^m}}}$
$ = \mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^n}}}{{1 - {x^m}}} = \mathop {\lim }\limits_{x \to 1} \frac{{(1 - x)({x^{n - 1}} + {x^{n - 2}} + ... + 1)}}{{(1 - x)({x^{m - 1}} + {x^{m - 2}} + ... + 1)}} = \frac{n}{m}.$
A. + ∞
B. - ∞
C. 5/2
D. 1
Chọn D
Ta có:$B = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} (\frac{\pi }{2} - x)\frac{{\sin x}}{{\cos {\rm{x}}}} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\frac{\pi }{2} - x}}{{\sin (\frac{\pi }{2} - x)}}.\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \sin x = 1$.
Ta có:$B = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} (\frac{\pi }{2} - x)\frac{{\sin x}}{{\cos {\rm{x}}}} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\frac{\pi }{2} - x}}{{\sin (\frac{\pi }{2} - x)}}.\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \sin x = 1$.
A. + ∞
B. - ∞
C. 5/2
D. 0
Chọn D
Ta có:$0 \le |{x^\alpha }\sin \frac{1}{x}| < {x^\alpha }$. Mà $\mathop {\lim }\limits_{x \to 0} {x^\alpha } = 0$
Nên theo nguyên lí kẹp$ \Rightarrow {A_{39}} = 0$.
Ta có:$0 \le |{x^\alpha }\sin \frac{1}{x}| < {x^\alpha }$. Mà $\mathop {\lim }\limits_{x \to 0} {x^\alpha } = 0$
Nên theo nguyên lí kẹp$ \Rightarrow {A_{39}} = 0$.
A. + ∞
B. - ∞
C. 5/2
D. 0
Chọn D
Trước hết ta có: $\sin x < x{\rm{ }}\forall x > 0$
Ta có: $\left| {\sin \sqrt {x + 1} - \sin \sqrt x } \right| = \left| {2{\rm{sin}}\frac{{\sqrt {x + 1} - \sqrt x }}{2}.\cos \frac{{\sqrt {x + 1} + \sqrt x }}{2}} \right|$$ < \frac{1}{{\sqrt {x + 1} + \sqrt x }}$
Mà $\mathop {\lim }\limits_{x \to + \infty } \frac{1}{{\sqrt {x + 1} + \sqrt x }} = 0$ nên$D = 0$.
Trước hết ta có: $\sin x < x{\rm{ }}\forall x > 0$
Ta có: $\left| {\sin \sqrt {x + 1} - \sin \sqrt x } \right| = \left| {2{\rm{sin}}\frac{{\sqrt {x + 1} - \sqrt x }}{2}.\cos \frac{{\sqrt {x + 1} + \sqrt x }}{2}} \right|$$ < \frac{1}{{\sqrt {x + 1} + \sqrt x }}$
Mà $\mathop {\lim }\limits_{x \to + \infty } \frac{1}{{\sqrt {x + 1} + \sqrt x }} = 0$ nên$D = 0$.
A. + ∞
B. - ∞
C. 7/11
D. 0
Chọn C
Ta có: $A = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{{7x}}{2}\sin \frac{x}{2}}}{{\sin \frac{{11x}}{2}\sin \frac{x}{2}}} = \frac{7}{{11}}$
Ta có: $A = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{{7x}}{2}\sin \frac{x}{2}}}{{\sin \frac{{11x}}{2}\sin \frac{x}{2}}} = \frac{7}{{11}}$
A. + ∞
B. - ∞
C. - 4/9
D. 0
Chọn C
Ta có \(B = \mathop {\lim }\limits_{x \to 0} \frac{{ - 2\sin 2x}}{{\sin 3x\left( {1 + \sqrt[3]{{1 + 2\sin 2x}} + \sqrt[3]{{{{(1 + 2\sin 2x)}^2}}}} \right)}} = - \frac{4}{9}\)
Ta có \(B = \mathop {\lim }\limits_{x \to 0} \frac{{ - 2\sin 2x}}{{\sin 3x\left( {1 + \sqrt[3]{{1 + 2\sin 2x}} + \sqrt[3]{{{{(1 + 2\sin 2x)}^2}}}} \right)}} = - \frac{4}{9}\)
A. + ∞
B. - ∞
C. \( - 96\)
D. 0
Chọn C
Ta có: \(C = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{{\sin }^2}2x}}{{{x^2}}}}}{{\frac{{\sqrt[3]{{\cos x}} - 1}}{{{x^2}}} + \frac{{1 - \sqrt[4]{{\cos x}}}}{{{x^2}}}}} = - 96\)
Ta có: \(C = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{{\sin }^2}2x}}{{{x^2}}}}}{{\frac{{\sqrt[3]{{\cos x}} - 1}}{{{x^2}}} + \frac{{1 - \sqrt[4]{{\cos x}}}}{{{x^2}}}}} = - 96\)
A. + ∞
B. - ∞
C. 16/81
D. 0
Chọn C
A. + ∞
B. - ∞
C. 5/2
D. 0
Chọn D
\(E = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{1 - \sin \left( {\frac{\pi }{2}\cos x} \right)}}{{\tan x}}}}{{\frac{{\sin (\tan x)}}{{\tan x}}}} = 0\)
\(E = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{1 - \sin \left( {\frac{\pi }{2}\cos x} \right)}}{{\tan x}}}}{{\frac{{\sin (\tan x)}}{{\tan x}}}} = 0\)
A. + ∞
B. - ∞
C. 5/2
D. 0
Chọn D
Ta có: \(0 \le \frac{{\left| {3\sin x + 2\cos x} \right|}}{{\sqrt {x + 1} + \sqrt x }} < \frac{1}{{\sqrt {x + 1} + \sqrt x }} \to 0\) khi \(x \to + \infty \)
Vậy \(F = 0\).
Ta có: \(0 \le \frac{{\left| {3\sin x + 2\cos x} \right|}}{{\sqrt {x + 1} + \sqrt x }} < \frac{1}{{\sqrt {x + 1} + \sqrt x }} \to 0\) khi \(x \to + \infty \)
Vậy \(F = 0\).
A. + ∞
B. - ∞
C. \(\frac{b}{{2n}} - \frac{a}{{2m}}\)
D. 0
Chọn C
Ta có: \(H = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sqrt[m]{{\cos ax}} - 1}}{{{x^2}}} + \frac{{1 - \sqrt[n]{{\cos bx}}}}{{{x^2}}}}}{{\frac{{{{\sin }^2}x}}{{{x^2}}}}} = \frac{b}{{2n}} - \frac{a}{{2m}}\)
Ta có: \(H = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sqrt[m]{{\cos ax}} - 1}}{{{x^2}}} + \frac{{1 - \sqrt[n]{{\cos bx}}}}{{{x^2}}}}}{{\frac{{{{\sin }^2}x}}{{{x^2}}}}} = \frac{b}{{2n}} - \frac{a}{{2m}}\)
A. + ∞
B. - ∞
C. \(\frac{a}{{2n}}\)
D. 0
Chọn C
Ta có: $1 - \sqrt[n]{{\cos ax}} = \frac{{1 - \cos ax}}{{1 + \sqrt[n]{{\cos ax}} + {{(\sqrt[n]{{\cos ax}})}^2} + ... + {{(\sqrt[n]{{\cos ax}})}^{n - 1}}}}$
$ \Rightarrow M = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos a{\rm{x}}}}{{{x^2}}}\mathop {\lim }\limits_{x \to 0} \frac{1}{{1 + \sqrt[n]{{\cos ax}} + {{(\sqrt[n]{{\cos ax}})}^2} + ... + {{(\sqrt[n]{{\cos ax}})}^{n - 1}}}}$ $ = \frac{a}{2}.\frac{1}{n} = \frac{a}{{2n}}$.
Ta có: $1 - \sqrt[n]{{\cos ax}} = \frac{{1 - \cos ax}}{{1 + \sqrt[n]{{\cos ax}} + {{(\sqrt[n]{{\cos ax}})}^2} + ... + {{(\sqrt[n]{{\cos ax}})}^{n - 1}}}}$
$ \Rightarrow M = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos a{\rm{x}}}}{{{x^2}}}\mathop {\lim }\limits_{x \to 0} \frac{1}{{1 + \sqrt[n]{{\cos ax}} + {{(\sqrt[n]{{\cos ax}})}^2} + ... + {{(\sqrt[n]{{\cos ax}})}^{n - 1}}}}$ $ = \frac{a}{2}.\frac{1}{n} = \frac{a}{{2n}}$.
A. + ∞
B. - ∞
C. 7/11
D. 0
Chọn C
Ta có: \(A = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{{7x}}{2}\sin \frac{x}{2}}}{{\sin \frac{{11x}}{2}\sin \frac{x}{2}}} = \frac{7}{{11}}\)
Ta có: \(A = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{{7x}}{2}\sin \frac{x}{2}}}{{\sin \frac{{11x}}{2}\sin \frac{x}{2}}} = \frac{7}{{11}}\)
A. + ∞
B. - ∞
C. - 4/9
D. 0
Chọn C
Ta có \(B = \mathop {\lim }\limits_{x \to 0} \frac{{ - 2\sin 2x}}{{\sin 3x\left( {1 + \sqrt[3]{{1 + 2\sin 2x}} + \sqrt[3]{{{{(1 + 2\sin 2x)}^2}}}} \right)}} = - \frac{4}{9}\)
Ta có \(B = \mathop {\lim }\limits_{x \to 0} \frac{{ - 2\sin 2x}}{{\sin 3x\left( {1 + \sqrt[3]{{1 + 2\sin 2x}} + \sqrt[3]{{{{(1 + 2\sin 2x)}^2}}}} \right)}} = - \frac{4}{9}\)
A. + ∞
B. - ∞
C. \( - 96\)
D. 0
Chọn C
Ta có: \(C = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{{\sin }^2}2x}}{{{x^2}}}}}{{\frac{{\sqrt[3]{{\cos x}} - 1}}{{{x^2}}} + \frac{{1 - \sqrt[4]{{\cos x}}}}{{{x^2}}}}} = - 96\)
Ta có: \(C = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{{\sin }^2}2x}}{{{x^2}}}}}{{\frac{{\sqrt[3]{{\cos x}} - 1}}{{{x^2}}} + \frac{{1 - \sqrt[4]{{\cos x}}}}{{{x^2}}}}} = - 96\)
A. $ + \infty $
B. - ∞
C. 16/81
D. 0
Chọn C
Ta có: \(D = \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{\sin 2x}}{{2x}}} \right)^4}.{\left( {\frac{{3x}}{{\sin 3x}}} \right)^4}.\frac{{16}}{{81}} = \frac{{16}}{{81}}\)
Ta có: \(D = \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{\sin 2x}}{{2x}}} \right)^4}.{\left( {\frac{{3x}}{{\sin 3x}}} \right)^4}.\frac{{16}}{{81}} = \frac{{16}}{{81}}\)
A. + ∞
B. - ∞
C. 1
D. 0
Chọn D
Ta có: \(E = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{1 - \sin \left( {\frac{\pi }{2}\cos x} \right)}}{{\tan x}}}}{{\frac{{\sin (\tan x)}}{{\tan x}}}}\)Mà \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin (\tan x)}}{{\tan x}} = 1\);
\(\mathop {\lim }\limits_{x \to 0} \frac{{1 - \sin \left( {\frac{\pi }{2}\cos x} \right)}}{{\tan x}} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left[ {\frac{\pi }{2}(1 - \cos x)} \right]}}{{\tan x}} = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}\left( {\frac{{\pi {{\sin }^2}\frac{x}{2}}}{2}} \right)}}{{\tan x}}\)
\( = \frac{\pi }{4}\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}\left( {\frac{{\pi {{\sin }^2}\frac{x}{2}}}{2}} \right)}}{{\frac{{\pi {{\sin }^2}\frac{x}{2}}}{2}}}\frac{{{{\sin }^2}\frac{x}{2}}}{{{{(\frac{x}{2})}^2}}}.x.\frac{x}{{\tan x}} = 0\)
Do đó: \(E = 0\).
Ta có: \(E = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{1 - \sin \left( {\frac{\pi }{2}\cos x} \right)}}{{\tan x}}}}{{\frac{{\sin (\tan x)}}{{\tan x}}}}\)Mà \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin (\tan x)}}{{\tan x}} = 1\);
\(\mathop {\lim }\limits_{x \to 0} \frac{{1 - \sin \left( {\frac{\pi }{2}\cos x} \right)}}{{\tan x}} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left[ {\frac{\pi }{2}(1 - \cos x)} \right]}}{{\tan x}} = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}\left( {\frac{{\pi {{\sin }^2}\frac{x}{2}}}{2}} \right)}}{{\tan x}}\)
\( = \frac{\pi }{4}\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}\left( {\frac{{\pi {{\sin }^2}\frac{x}{2}}}{2}} \right)}}{{\frac{{\pi {{\sin }^2}\frac{x}{2}}}{2}}}\frac{{{{\sin }^2}\frac{x}{2}}}{{{{(\frac{x}{2})}^2}}}.x.\frac{x}{{\tan x}} = 0\)
Do đó: \(E = 0\).
A. + ∞
B. - ∞
C. 5/2
D. 0
Chọn D
Ta có: \(0 \le \frac{{\left| {3\sin x + 2\cos x} \right|}}{{\sqrt {x + 1} + \sqrt x }} < \frac{1}{{\sqrt {x + 1} + \sqrt x }} \to 0\) khi \(x \to + \infty \)
Vậy \(F = 0\).
Ta có: \(0 \le \frac{{\left| {3\sin x + 2\cos x} \right|}}{{\sqrt {x + 1} + \sqrt x }} < \frac{1}{{\sqrt {x + 1} + \sqrt x }} \to 0\) khi \(x \to + \infty \)
Vậy \(F = 0\).
A. + ∞
B. - ∞
C. \( - \frac{1}{4}\)
D. 0
Chọn C
Ta có: \(M = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sqrt[3]{{3x + 1}} - \sqrt {2x + 1} }}{{{x^2}}}}}{{\frac{{1 - \cos 2x}}{{{x^2}}}}} = \frac{{ - \frac{1}{2}}}{2} = - \frac{1}{4}\).
Ta có: \(M = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sqrt[3]{{3x + 1}} - \sqrt {2x + 1} }}{{{x^2}}}}}{{\frac{{1 - \cos 2x}}{{{x^2}}}}} = \frac{{ - \frac{1}{2}}}{2} = - \frac{1}{4}\).
A. - ∞.
B. 0.
C. \(3\).
D. + ∞.
Chọn B.
Vậy$\mathop {\lim }\limits_{x \to + \infty } \frac{{3x - 5\sin 2x + {{\cos }^2}x}}{{{x^2} + 2}} = 0$.
Vậy$\mathop {\lim }\limits_{x \to + \infty } \frac{{3x - 5\sin 2x + {{\cos }^2}x}}{{{x^2} + 2}} = 0$.