TÍNH GIỚI HẠN DẠNG VÔ ĐỊNH $\frac{0}{0}$
1. L = \(\mathop {\lim }\limits_{x \to {x_0}} \frac{{P(x)}}{{Q(x)}}\) với P(x), Q(x) là các đa thức và P(x$_0$) = Q(x$_0$) = 0
Phân tích cả tử và mẫu thành nhân tử và rút gọn.
Chú ý:
Sử dụng các hằng đẳng thức để nhân lượng liên hợp ở tử và mẫu.
Các lượng liên hợp:
Giả sử: P(x) = \(\sqrt[m]{{u(x)}} - \sqrt[n]{{v(x)}}\,\,v\^o \`u i\,\,\sqrt[m]{{u({x_0})}} = \sqrt[n]{{v({x_0})}} = a\).
Ta phân tích P(x) = \(\left( {\sqrt[m]{{u(x)}} - a} \right) + \left( {a - \sqrt[n]{{v(x)}}} \right)\).
Trong nhiều trường hợp việc phân tích như trên không đi đến kết quả ta phải phân tích như sau: $\sqrt[n]{{u(x)}} - \sqrt[m]{{v(x)}} = (\sqrt[n]{{u(x)}} - m(x)) - (\sqrt[m]{{v(x)}} - m(x))$, trong đó $m(x) \to c$.
1. L = \(\mathop {\lim }\limits_{x \to {x_0}} \frac{{P(x)}}{{Q(x)}}\) với P(x), Q(x) là các đa thức và P(x$_0$) = Q(x$_0$) = 0
Phân tích cả tử và mẫu thành nhân tử và rút gọn.
Chú ý:
- Nếu tam thức bậc hai $a{x^2} + b{\rm{x + c}}$ có hai nghiệm ${x_1},{x_2}$ thì ta luôn có sự phân tích$a{x^2} + bx + c = a(x - {x_1})(x - {x_2})$.
- ${a^n} - {b^n} = (a - b)({a^{n - 1}} + {a^{n - 2}}b + ... + a{b^{n - 2}} + {b^{n - 1}})$
Sử dụng các hằng đẳng thức để nhân lượng liên hợp ở tử và mẫu.
Các lượng liên hợp:
- $(\sqrt a - \sqrt b )(\sqrt a + \sqrt b ) = a - b$
- $(\sqrt[3]{a} \pm \sqrt[3]{b})(\sqrt[3]{{{a^2}}} \mp \sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}) = a - b$
- $(\sqrt[n]{a} - \sqrt[n]{b})(\sqrt[n]{{{a^{n - 1}}}} + \sqrt[n]{{{a^{n - 2}}b}} + ... + \sqrt[n]{{{b^{n - 1}}}}) = a - b$
Giả sử: P(x) = \(\sqrt[m]{{u(x)}} - \sqrt[n]{{v(x)}}\,\,v\^o \`u i\,\,\sqrt[m]{{u({x_0})}} = \sqrt[n]{{v({x_0})}} = a\).
Ta phân tích P(x) = \(\left( {\sqrt[m]{{u(x)}} - a} \right) + \left( {a - \sqrt[n]{{v(x)}}} \right)\).
Trong nhiều trường hợp việc phân tích như trên không đi đến kết quả ta phải phân tích như sau: $\sqrt[n]{{u(x)}} - \sqrt[m]{{v(x)}} = (\sqrt[n]{{u(x)}} - m(x)) - (\sqrt[m]{{v(x)}} - m(x))$, trong đó $m(x) \to c$.
Câu 1. Chọn kết quả đúng trong các kết quả sau của \(\mathop {\lim }\limits_{x \to - 1} \frac{{{x^2} + 2x + 1}}{{2{x^3} + 2}}\) là:
A. - ∞.
B. 0.
C. 1/2.
D. + ∞.
Chọn B.
Cách 1: \(\mathop {\lim }\limits_{x \to - 1} \frac{{{x^2} + 2x + 1}}{{2{x^3} + 2}}\)\( = \mathop {\lim }\limits_{x \to - 1} \frac{{{{\left( {x + 1} \right)}^2}}}{{2\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\)\( = \mathop {\lim }\limits_{x \to - 1} \frac{{x + 1}}{{2\left( {{x^2} - x + 1} \right)}} = 0\)
Cách 2: Bấm máy tính như sau: \(\frac{{{x^2} + 2x + 1}}{{2{x^3} + 2}}\) + CACL + \(x = - 1 + {10^{ - 9}}\) và so đáp án.
Cách 3: Dùng chức lim của máy VNCALL 570ES Plus: ${\left. {\lim \frac{{{x^2} + 2x + 1}}{{2{x^3} + 2}}} \right|_{x \to - 1 + {{10}^{ - 9}}}}$ và so đáp án.
Cách 1: \(\mathop {\lim }\limits_{x \to - 1} \frac{{{x^2} + 2x + 1}}{{2{x^3} + 2}}\)\( = \mathop {\lim }\limits_{x \to - 1} \frac{{{{\left( {x + 1} \right)}^2}}}{{2\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\)\( = \mathop {\lim }\limits_{x \to - 1} \frac{{x + 1}}{{2\left( {{x^2} - x + 1} \right)}} = 0\)
Cách 2: Bấm máy tính như sau: \(\frac{{{x^2} + 2x + 1}}{{2{x^3} + 2}}\) + CACL + \(x = - 1 + {10^{ - 9}}\) và so đáp án.
Cách 3: Dùng chức lim của máy VNCALL 570ES Plus: ${\left. {\lim \frac{{{x^2} + 2x + 1}}{{2{x^3} + 2}}} \right|_{x \to - 1 + {{10}^{ - 9}}}}$ và so đáp án.
A. + ∞
B. - 11/4
C. 3/2
D. 1
Chọn C
Ta có: $A = \mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 3{x^2} + 2}}{{{x^2} - 4x + 3}} = \mathop {\lim }\limits_{x \to 1} \frac{{(x - 1)({x^2} - 2x - 2)}}{{(x - 1)(x - 3)}}$$ = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 2x - 2}}{{x - 3}} = \frac{3}{2}$.
Ta có: $A = \mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 3{x^2} + 2}}{{{x^2} - 4x + 3}} = \mathop {\lim }\limits_{x \to 1} \frac{{(x - 1)({x^2} - 2x - 2)}}{{(x - 1)(x - 3)}}$$ = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 2x - 2}}{{x - 3}} = \frac{3}{2}$.
A. + ∞
B. - ∞
C. - 1/6
D. 1
Chọn D
Ta có: $B = \mathop {\lim }\limits_{x \to 2} \frac{{{x^4} - 5{x^2} + 4}}{{{x^3} - 8}} = \mathop {\lim }\limits_{x \to 2} \frac{{({x^2} - 1)({x^2} - 4)}}{{{x^3} - {2^3}}}$$ = \mathop {\lim }\limits_{x \to 2} \frac{{({x^2} - 1)(x - 2)(x + 2)}}{{(x - 2)({x^2} + 2x + 4)}}$$ = \mathop {\lim }\limits_{x \to 2} \frac{{({x^2} - 1)(x + 2)}}{{{x^2} + 2x + 4}} = 1$.
Ta có: $B = \mathop {\lim }\limits_{x \to 2} \frac{{{x^4} - 5{x^2} + 4}}{{{x^3} - 8}} = \mathop {\lim }\limits_{x \to 2} \frac{{({x^2} - 1)({x^2} - 4)}}{{{x^3} - {2^3}}}$$ = \mathop {\lim }\limits_{x \to 2} \frac{{({x^2} - 1)(x - 2)(x + 2)}}{{(x - 2)({x^2} + 2x + 4)}}$$ = \mathop {\lim }\limits_{x \to 2} \frac{{({x^2} - 1)(x + 2)}}{{{x^2} + 2x + 4}} = 1$.
A. + ∞
B. - ∞
C. - 1/6
D. 25
Chọn D
Ta có: $C = \mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + 3x)}^3} - {{(1 - 4x)}^4}}}{x}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + 3x)}^3} - 1}}{x} - \mathop {\lim }\limits_{x \to 0} \frac{{{{(1 - 4x)}^4} - 1}}{x}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{3x[{{(1 + 3x)}^2} + (1 + 3x) + 1]}}{x} - \mathop {\lim }\limits_{x \to 0} \frac{{ - 4x(2 - 4x)[{{(1 - 4x)}^2} + 1]}}{x}$
$ = \mathop {\lim }\limits_{x \to 0} 3[{(1 + 3x)^2} + (1 + 3x) + 1] + \mathop {\lim }\limits_{x \to 0} 4(2 - 4x)[{(1 - 4x)^2} + 1] = 25$
Ta có: $C = \mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + 3x)}^3} - {{(1 - 4x)}^4}}}{x}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + 3x)}^3} - 1}}{x} - \mathop {\lim }\limits_{x \to 0} \frac{{{{(1 - 4x)}^4} - 1}}{x}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{3x[{{(1 + 3x)}^2} + (1 + 3x) + 1]}}{x} - \mathop {\lim }\limits_{x \to 0} \frac{{ - 4x(2 - 4x)[{{(1 - 4x)}^2} + 1]}}{x}$
$ = \mathop {\lim }\limits_{x \to 0} 3[{(1 + 3x)^2} + (1 + 3x) + 1] + \mathop {\lim }\limits_{x \to 0} 4(2 - 4x)[{(1 - 4x)^2} + 1] = 25$
A. - ∞.
B. $0.$.
C. $\sqrt 6 .$.
D. + ∞
Chọn B
$\mathop {\lim }\limits_{x \to {3^ + }} \frac{{x - 3}}{{\sqrt {{x^2} - 9} }} = \mathop {\lim }\limits_{x \to {3^ + }} \frac{{\sqrt {{{\left( {x - 3} \right)}^2}} }}{{\sqrt {\left( {x - 3} \right)\left( {x + 3} \right)} }}$.
$ = \mathop {\lim }\limits_{x \to {3^ + }} \frac{{\sqrt {\left( {x - 3} \right)} }}{{\sqrt {\left( {x + 3} \right)} }} = 0$.
$\mathop {\lim }\limits_{x \to {3^ + }} \frac{{x - 3}}{{\sqrt {{x^2} - 9} }} = \mathop {\lim }\limits_{x \to {3^ + }} \frac{{\sqrt {{{\left( {x - 3} \right)}^2}} }}{{\sqrt {\left( {x - 3} \right)\left( {x + 3} \right)} }}$.
$ = \mathop {\lim }\limits_{x \to {3^ + }} \frac{{\sqrt {\left( {x - 3} \right)} }}{{\sqrt {\left( {x + 3} \right)} }} = 0$.
A. + ∞
B. - ∞
C. - 1/6
D. 6
Chọn D
Ta có:$D = \mathop {\lim }\limits_{x \to 0} \frac{{(1 + x)(1 + 2x)(1 + 3x) - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{6{x^3} + 11{x^2} + 6x}}{x} = 6$.
Ta có:$D = \mathop {\lim }\limits_{x \to 0} \frac{{(1 + x)(1 + 2x)(1 + 3x) - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{6{x^3} + 11{x^2} + 6x}}{x} = 6$.
A. + ∞
B. - ∞
C. \(\frac{n}{m}\)
D. \(m - n\)
Chọn C
Ta có: $A = \mathop {\lim }\limits_{x \to 0} \frac{{(x - 1)({x^{n - 1}} + {x^{n - 2}} + ... + x + 1)}}{{(x - 1)({x^{m - 1}} + {x^{m - 2}} + ... + x + 1)}}$$ = \mathop {\lim }\limits_{x \to 0} \frac{{{x^{n - 1}} + {x^{n - 2}} + ... + x + 1}}{{{x^{m - 1}} + {x^{m - 2}} + ... + x + 1}} = \frac{n}{m}$.
Ta có: $A = \mathop {\lim }\limits_{x \to 0} \frac{{(x - 1)({x^{n - 1}} + {x^{n - 2}} + ... + x + 1)}}{{(x - 1)({x^{m - 1}} + {x^{m - 2}} + ... + x + 1)}}$$ = \mathop {\lim }\limits_{x \to 0} \frac{{{x^{n - 1}} + {x^{n - 2}} + ... + x + 1}}{{{x^{m - 1}} + {x^{m - 2}} + ... + x + 1}} = \frac{n}{m}$.
A. + ∞
B. - ∞
C. \(\frac{a}{n}\)
D. \(1 - \frac{n}{a}\)
Chọn C
Cách 1: Nhân liên hợp
Ta có:
$B = \mathop {\lim }\limits_{x \to 0} \frac{{(\sqrt[n]{{1 + ax}} - 1)(\sqrt[n]{{{{(1 + ax)}^{n - 1}}}} + \sqrt[n]{{{{(1 + ax)}^{n - 2}}}} + ... + \sqrt[n]{{1 + ax}} + 1)}}{{x(\sqrt[n]{{{{(1 + ax)}^{n - 1}}}} + \sqrt[n]{{{{(1 + ax)}^{n - 2}}}} + ... + \sqrt[n]{{1 + ax}} + 1)}}$
$B = \mathop {\lim }\limits_{x \to 0} \frac{a}{{\sqrt[n]{{{{(1 + ax)}^{n - 1}}}} + \sqrt[n]{{{{(1 + ax)}^{n - 2}}}} + ... + \sqrt[n]{{1 + ax}} + 1}} = \frac{a}{n}$.
Cách 2: Đặt ẩn phụ
Đặt $t = \sqrt[n]{{1 + ax}} \Rightarrow x = \frac{{{t^n} - 1}}{a}$ và $x \to 0 \Leftrightarrow t \to 1$
$ \Rightarrow B = a\mathop {\lim }\limits_{t \to 1} \frac{{t - 1}}{{{t^n} - 1}} = a\mathop {\lim }\limits_{t \to 1} \frac{{t - 1}}{{(t - 1)({t^{n - 1}} + {t^n} + ... + t + 1)}} = \frac{a}{n}$.
Cách 1: Nhân liên hợp
Ta có:
$B = \mathop {\lim }\limits_{x \to 0} \frac{{(\sqrt[n]{{1 + ax}} - 1)(\sqrt[n]{{{{(1 + ax)}^{n - 1}}}} + \sqrt[n]{{{{(1 + ax)}^{n - 2}}}} + ... + \sqrt[n]{{1 + ax}} + 1)}}{{x(\sqrt[n]{{{{(1 + ax)}^{n - 1}}}} + \sqrt[n]{{{{(1 + ax)}^{n - 2}}}} + ... + \sqrt[n]{{1 + ax}} + 1)}}$
$B = \mathop {\lim }\limits_{x \to 0} \frac{a}{{\sqrt[n]{{{{(1 + ax)}^{n - 1}}}} + \sqrt[n]{{{{(1 + ax)}^{n - 2}}}} + ... + \sqrt[n]{{1 + ax}} + 1}} = \frac{a}{n}$.
Cách 2: Đặt ẩn phụ
Đặt $t = \sqrt[n]{{1 + ax}} \Rightarrow x = \frac{{{t^n} - 1}}{a}$ và $x \to 0 \Leftrightarrow t \to 1$
$ \Rightarrow B = a\mathop {\lim }\limits_{t \to 1} \frac{{t - 1}}{{{t^n} - 1}} = a\mathop {\lim }\limits_{t \to 1} \frac{{t - 1}}{{(t - 1)({t^{n - 1}} + {t^n} + ... + t + 1)}} = \frac{a}{n}$.
A. $ + \infty $
B. - ∞
C. \(\frac{{am}}{{bn}}\)
D. \(1 + \frac{{am}}{{bn}}\)
Chọn C
Áp dụng bài toán trên ta có:
$A = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{1 + ax}} - 1}}{x}.\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt[m]{{1 + bx}} - 1}} = \frac{a}{n}.\frac{m}{b} = \frac{{am}}{{bn}}$.
Áp dụng bài toán trên ta có:
$A = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{1 + ax}} - 1}}{x}.\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt[m]{{1 + bx}} - 1}} = \frac{a}{n}.\frac{m}{b} = \frac{{am}}{{bn}}$.
A. + ∞
B. - ∞
C. $B = \frac{\gamma }{4} - \frac{\beta }{3} + \frac{\alpha }{2}$
D. $B = \frac{\gamma }{4} + \frac{\beta }{3} + \frac{\alpha }{2}$
Chọn C
Ta có: $\sqrt {1 + \alpha x} \sqrt[3]{{1 + \beta x}}\sqrt[4]{{1 + \gamma x}} - 1 = $
$ = \sqrt {1 + \alpha x} \sqrt[3]{{1 + \beta x}}(\sqrt[4]{{1 + \gamma x}} - 1) + \sqrt {1 + \alpha x} ((\sqrt[3]{{1 + \beta x}} - 1) + (\sqrt {1 + \alpha x} - 1)$
$B = \mathop {\lim }\limits_{x \to 0} (\sqrt {1 + \alpha x} \sqrt[3]{{1 + \beta x}})\frac{{\sqrt[4]{{1 + \gamma x}} - 1}}{x} + \mathop {\lim }\limits_{x \to 0} \sqrt {1 + \alpha x} \frac{{\sqrt[3]{{1 + \beta x}} - 1}}{x}$$ + \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + \alpha x} - 1}}{x}$
Ta có: $\sqrt {1 + \alpha x} \sqrt[3]{{1 + \beta x}}\sqrt[4]{{1 + \gamma x}} - 1 = $
$ = \sqrt {1 + \alpha x} \sqrt[3]{{1 + \beta x}}(\sqrt[4]{{1 + \gamma x}} - 1) + \sqrt {1 + \alpha x} ((\sqrt[3]{{1 + \beta x}} - 1) + (\sqrt {1 + \alpha x} - 1)$
$B = \mathop {\lim }\limits_{x \to 0} (\sqrt {1 + \alpha x} \sqrt[3]{{1 + \beta x}})\frac{{\sqrt[4]{{1 + \gamma x}} - 1}}{x} + \mathop {\lim }\limits_{x \to 0} \sqrt {1 + \alpha x} \frac{{\sqrt[3]{{1 + \beta x}} - 1}}{x}$$ + \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + \alpha x} - 1}}{x}$
A. + ∞
B. - ∞
C. 1/3
D. 1
Chọn C
Ta có: \(A = \mathop {\lim }\limits_{x \to 2} \frac{{(x - 2)(2x - 1)}}{{(x - 2)({x^2} + 2x + 1)}} = \frac{1}{3}\)
Ta có: \(A = \mathop {\lim }\limits_{x \to 2} \frac{{(x - 2)(2x - 1)}}{{(x - 2)({x^2} + 2x + 1)}} = \frac{1}{3}\)
A. + ∞
B. - ∞
C. 1/5
D. 1
Chọn C
Ta có: \(B = \mathop {\lim }\limits_{x \to 1} \frac{{(x - 1)({x^3} + {x^2} + x - 2)}}{{(x - 1)({x^2} + x + 3)}} = \frac{1}{5}\)
Ta có: \(B = \mathop {\lim }\limits_{x \to 1} \frac{{(x - 1)({x^3} + {x^2} + x - 2)}}{{(x - 1)({x^2} + x + 3)}} = \frac{1}{5}\)
A. + ∞
B. - ∞
C. - 1/3
D. 1
Chọn C
Ta có: \(C = \mathop {\lim }\limits_{x \to 3} \frac{{ - (x - 3)(x + 1)}}{{(x - 3)(x - 1)\left( {\sqrt {2x + 3} + x} \right)}} = \frac{{ - 1}}{3}\)
Ta có: \(C = \mathop {\lim }\limits_{x \to 3} \frac{{ - (x - 3)(x + 1)}}{{(x - 3)(x - 1)\left( {\sqrt {2x + 3} + x} \right)}} = \frac{{ - 1}}{3}\)
A. + ∞
B. - ∞
C. 2/3
D. 1
Chọn C
Ta có: \(D = \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {\sqrt[4]{{{{(2x + 1)}^3}}} + \sqrt[4]{{{{(2x + 1)}^2}}} + \sqrt[4]{{2x + 1}} + 1} \right)}}{{2x\left( {\sqrt[3]{{{{(x + 1)}^2}}} + \sqrt[3]{{x + 1}} + 1} \right)}} = \frac{2}{3}\)
Ta có: \(D = \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {\sqrt[4]{{{{(2x + 1)}^3}}} + \sqrt[4]{{{{(2x + 1)}^2}}} + \sqrt[4]{{2x + 1}} + 1} \right)}}{{2x\left( {\sqrt[3]{{{{(x + 1)}^2}}} + \sqrt[3]{{x + 1}} + 1} \right)}} = \frac{2}{3}\)
A. + ∞
B. - ∞
C. \(\frac{{ - 8}}{{27}}\)
D. 1
Chọn C
Ta có: $E = \mathop {\lim }\limits_{x \to 7} \frac{{\sqrt[3]{{4x - 1}} - \sqrt {x + 2} }}{{\sqrt[4]{{2x + 2}} - 2}} = \mathop {\lim }\limits_{x \to 7} \frac{{\sqrt[3]{{4x - 1}} - 3}}{{\sqrt[4]{{2x + 2}} - 2}} - \mathop {\lim }\limits_{x \to 7} \frac{{\sqrt {x + 2} - 3}}{{\sqrt[4]{{2x + 2}} - 2}} = A - B$
$A = \mathop {\lim }\limits_{x \to 7} \frac{{\sqrt[3]{{4x - 1}} - 3}}{{\sqrt[4]{{2x + 2}} - 2}} = \mathop {\lim }\limits_{x \to 7} \frac{{2\left( {\sqrt[4]{{2x + 2}} + 2} \right)\left( {\sqrt[4]{{{{\left( {2x + 2} \right)}^2}}} + 4} \right)}}{{\left( {\sqrt[3]{{{{\left( {4x - 1} \right)}^2}}} + 3\sqrt[3]{{4x - 1}} + 9} \right)}} = \frac{{64}}{{27}}$
$B = \mathop {\lim }\limits_{x \to 7} \frac{{\sqrt {x + 2} - 3}}{{\sqrt[4]{{2x + 2}} - 2}} = \mathop {\lim }\limits_{x \to 7} \frac{{\left( {\sqrt[4]{{2x + 2}} + 2} \right)\left( {\sqrt[4]{{{{\left( {2x + 2} \right)}^2}}} + 4} \right)}}{{2\left( {\sqrt {x + 2} + 3} \right)}} = \frac{8}{3}$
$E = A - B = \frac{{64}}{{27}} - \frac{8}{3} = \frac{{ - 8}}{{27}}$
Ta có: $E = \mathop {\lim }\limits_{x \to 7} \frac{{\sqrt[3]{{4x - 1}} - \sqrt {x + 2} }}{{\sqrt[4]{{2x + 2}} - 2}} = \mathop {\lim }\limits_{x \to 7} \frac{{\sqrt[3]{{4x - 1}} - 3}}{{\sqrt[4]{{2x + 2}} - 2}} - \mathop {\lim }\limits_{x \to 7} \frac{{\sqrt {x + 2} - 3}}{{\sqrt[4]{{2x + 2}} - 2}} = A - B$
$A = \mathop {\lim }\limits_{x \to 7} \frac{{\sqrt[3]{{4x - 1}} - 3}}{{\sqrt[4]{{2x + 2}} - 2}} = \mathop {\lim }\limits_{x \to 7} \frac{{2\left( {\sqrt[4]{{2x + 2}} + 2} \right)\left( {\sqrt[4]{{{{\left( {2x + 2} \right)}^2}}} + 4} \right)}}{{\left( {\sqrt[3]{{{{\left( {4x - 1} \right)}^2}}} + 3\sqrt[3]{{4x - 1}} + 9} \right)}} = \frac{{64}}{{27}}$
$B = \mathop {\lim }\limits_{x \to 7} \frac{{\sqrt {x + 2} - 3}}{{\sqrt[4]{{2x + 2}} - 2}} = \mathop {\lim }\limits_{x \to 7} \frac{{\left( {\sqrt[4]{{2x + 2}} + 2} \right)\left( {\sqrt[4]{{{{\left( {2x + 2} \right)}^2}}} + 4} \right)}}{{2\left( {\sqrt {x + 2} + 3} \right)}} = \frac{8}{3}$
$E = A - B = \frac{{64}}{{27}} - \frac{8}{3} = \frac{{ - 8}}{{27}}$
A. + ∞
B. - ∞
C. \(\frac{9}{2}\)
D. 1
Chọn C
A. + ∞
B. - ∞
C. 1/3
D. 0
Chọn D
Ta có: \(M = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {4x + 1} - (2x + 1)}}{{{x^2}}} - \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{1 + 6x}} - (2x + 1)}}{{{x^2}}} = 0\)
Ta có: \(M = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {4x + 1} - (2x + 1)}}{{{x^2}}} - \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{1 + 6x}} - (2x + 1)}}{{{x^2}}} = 0\)
A. + ∞
B. - ∞
C. \(\frac{a}{m} - \frac{b}{n}\)
D. \(\frac{a}{m} + \frac{b}{n}\)
Chọn C
Ta có: \(N = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[m]{{1 + ax}} - 1}}{x} - \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{1 + bx}} - 1}}{x} = \frac{a}{m} - \frac{b}{n}\)
Ta có: \(N = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[m]{{1 + ax}} - 1}}{x} - \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[n]{{1 + bx}} - 1}}{x} = \frac{a}{m} - \frac{b}{n}\)
A. + ∞
B. - ∞
C. \(\frac{a}{m} - \frac{b}{n}\)
D. \(\frac{a}{m} + \frac{b}{n}\)
Chọn D
Ta có: \(G = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[m]{{1 + ax}}\left( {\sqrt[n]{{1 + bx}} - 1} \right)}}{x} + \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[m]{{1 + ax}} - 1}}{x} = \frac{b}{n} + \frac{a}{m}\)
Ta có: \(G = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[m]{{1 + ax}}\left( {\sqrt[n]{{1 + bx}} - 1} \right)}}{x} + \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[m]{{1 + ax}} - 1}}{x} = \frac{b}{n} + \frac{a}{m}\)
A. + ∞
B. - ∞
C. $V = \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {1 + mx} \right)}^n} - {{\left( {1 + nx} \right)}^m}}}{{{x^2}}}$
D. \(\frac{{mn\left( {n + m} \right)}}{2}\)
Chọn C
Ta có: \(V = \mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + nx)}^m} - (1 + mnx)}}{{{x^2}}} - \mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + mx)}^n} - (1 + mnx)}}{{{x^2}}}\)\( = \frac{{mn(n - m)}}{2}\).
Ta có: \(V = \mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + nx)}^m} - (1 + mnx)}}{{{x^2}}} - \mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + mx)}^n} - (1 + mnx)}}{{{x^2}}}\)\( = \frac{{mn(n - m)}}{2}\).
A. + ∞
B. - ∞
C. \(\frac{1}{{n!}}\)
D. 0
Chọn C
Ta có: \(K = \mathop {\lim }\limits_{x \to 1} \frac{1}{{(1 + \sqrt x )(\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1)...(\sqrt[n]{{{x^{n - 1}}}} + ... + 1)}} = \frac{1}{{n!}}\).
Ta có: \(K = \mathop {\lim }\limits_{x \to 1} \frac{1}{{(1 + \sqrt x )(\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1)...(\sqrt[n]{{{x^{n - 1}}}} + ... + 1)}} = \frac{1}{{n!}}\).
A. + ∞
B. - ∞
C. \(2n\)
D. 0
Chọn C
\(L = \mathop {\lim }\limits_{x \to 0} \frac{{\left[ {{{\left( {\sqrt {1 + {x^2}} + x} \right)}^n} - 1} \right]\left[ {{{\left( {\sqrt {1 + {x^2}} + x} \right)}^n} + 1} \right]}}{{x{{\left( {\sqrt {1 + {x^2}} + x} \right)}^n}}} = 2n\).
\(L = \mathop {\lim }\limits_{x \to 0} \frac{{\left[ {{{\left( {\sqrt {1 + {x^2}} + x} \right)}^n} - 1} \right]\left[ {{{\left( {\sqrt {1 + {x^2}} + x} \right)}^n} + 1} \right]}}{{x{{\left( {\sqrt {1 + {x^2}} + x} \right)}^n}}} = 2n\).
A. + ∞
B. - ∞
C. 1/4
D. 0
Chọn C
Ta có: \(A = \mathop {\lim }\limits_{x \to 2} \frac{{(2x - 1)(x - 2)}}{{(x - 2)({x^2} + 2x + 4)}} = \frac{1}{4}\)
Ta có: \(A = \mathop {\lim }\limits_{x \to 2} \frac{{(2x - 1)(x - 2)}}{{(x - 2)({x^2} + 2x + 4)}} = \frac{1}{4}\)
A. + ∞
B. $ - \infty $
C. \( - \frac{2}{5}\)
D. 0
Chọn C
Ta có: $B = \mathop {\lim }\limits_{x \to 1} \frac{{({x^2} - 1)({x^2} - 2)}}{{(x - 1)({x^2} + x + 3)}} = - \frac{2}{5}$
Ta có: $B = \mathop {\lim }\limits_{x \to 1} \frac{{({x^2} - 1)({x^2} - 2)}}{{(x - 1)({x^2} + x + 3)}} = - \frac{2}{5}$
A. + ∞
B. - ∞
C. \(\frac{1}{6}\)
D. 0
Chọn C
Ta có: \(C = \mathop {\lim }\limits_{x \to 3} \frac{{2(x - 3)}}{{(x - 1)(x - 3)\left( {\sqrt {2x + 3} + 3} \right)}} = \frac{1}{6}\)
Ta có: \(C = \mathop {\lim }\limits_{x \to 3} \frac{{2(x - 3)}}{{(x - 1)(x - 3)\left( {\sqrt {2x + 3} + 3} \right)}} = \frac{1}{6}\)
A. + ∞
B. - ∞
C. 1/3
D. 0
Chọn C
Ta có: \(D = \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {\sqrt {2x + 1} + 1} \right)}}{{2x\left[ {\sqrt[3]{{{{(x + 1)}^2}}} + \sqrt[3]{{x + 1}} + 1} \right]}} = \frac{1}{3}\)
Ta có: \(D = \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {\sqrt {2x + 1} + 1} \right)}}{{2x\left[ {\sqrt[3]{{{{(x + 1)}^2}}} + \sqrt[3]{{x + 1}} + 1} \right]}} = \frac{1}{3}\)
A. + ∞
B. - ∞
C. \(\frac{9}{n}\)
D. 0
Chọn C
Đặt \(y = \sqrt[n]{{(2x + 1)(3x + 1)(4x + 1)}} \Rightarrow y \to 1\) khi x → 0
Và: \(\mathop {\lim }\limits_{x \to 0} \frac{{{y^n} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{(2x + 1)(3x + 1)(4x + 1) - 1}}{x} = 9\)
Do đó: \(F = \mathop {\lim }\limits_{x \to 0} \frac{{{y^n} - 1}}{{x\left( {{y^{n - 1}} + {y^{n - 2}} + ... + y + 1} \right)}} = \frac{9}{n}\)
Đặt \(y = \sqrt[n]{{(2x + 1)(3x + 1)(4x + 1)}} \Rightarrow y \to 1\) khi x → 0
Và: \(\mathop {\lim }\limits_{x \to 0} \frac{{{y^n} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{(2x + 1)(3x + 1)(4x + 1) - 1}}{x} = 9\)
Do đó: \(F = \mathop {\lim }\limits_{x \to 0} \frac{{{y^n} - 1}}{{x\left( {{y^{n - 1}} + {y^{n - 2}} + ... + y + 1} \right)}} = \frac{9}{n}\)
A. + ∞
B. - ∞
C. \(\frac{4}{9}\)
D. 0
Chọn C
Ta có: \(M = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 4x} - \sqrt[3]{{1 + 6x}}}}{{{x^2}}}.\frac{{{x^2}}}{{1 - \cos 3x}} = 2.\frac{2}{9} = \frac{4}{9}\).
Ta có: \(M = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 4x} - \sqrt[3]{{1 + 6x}}}}{{{x^2}}}.\frac{{{x^2}}}{{1 - \cos 3x}} = 2.\frac{2}{9} = \frac{4}{9}\).
A. + ∞
B. - ∞
C. \(\frac{{2\left( {an - bm} \right)}}{{mn}}\)
D. 0
Chọn C
Ta có: \(N = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sqrt[m]{{1 + ax}} - 1}}{x} - \frac{{\sqrt[n]{{1 + bx}} - 1}}{x}} \right).\frac{x}{{\sqrt {1 + x} - 1}}\)\( = \left( {\frac{a}{m} - \frac{b}{n}} \right).2 = \frac{{2(an - bm)}}{{mn}}\).
Ta có: \(N = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sqrt[m]{{1 + ax}} - 1}}{x} - \frac{{\sqrt[n]{{1 + bx}} - 1}}{x}} \right).\frac{x}{{\sqrt {1 + x} - 1}}\)\( = \left( {\frac{a}{m} - \frac{b}{n}} \right).2 = \frac{{2(an - bm)}}{{mn}}\).
A. + ∞
B. - ∞
C. \(\frac{{2\left( {an - bm} \right)}}{{mn}}\)
D. \(mn\left( {n - m} \right)\)
Chọn D
Ta có: \(V = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{{\left( {1 + mx} \right)}^n} - 1}}{{{x^2}}} - \frac{{{{(1 + nx)}^m} - 1}}{{{x^2}}}} \right]\frac{{{x^2}}}{{\sqrt {1 + 2x} - \sqrt[3]{{1 + 3x}}}}\)\( = \frac{{mn(n - m)}}{2}.2 = mn(n - m)\).
Ta có: \(V = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{{\left( {1 + mx} \right)}^n} - 1}}{{{x^2}}} - \frac{{{{(1 + nx)}^m} - 1}}{{{x^2}}}} \right]\frac{{{x^2}}}{{\sqrt {1 + 2x} - \sqrt[3]{{1 + 3x}}}}\)\( = \frac{{mn(n - m)}}{2}.2 = mn(n - m)\).
A. + ∞
B. - ∞
C. \(\frac{1}{{n!}}\)
D. 0
Chọn C
Ta có: \(K = \mathop {\lim }\limits_{x \to 1} \frac{1}{{(1 + \sqrt x )(\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1)...(\sqrt[n]{{{x^{n - 1}}}} + ... + 1)}} = \frac{1}{{n!}}\).
Ta có: \(K = \mathop {\lim }\limits_{x \to 1} \frac{1}{{(1 + \sqrt x )(\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1)...(\sqrt[n]{{{x^{n - 1}}}} + ... + 1)}} = \frac{1}{{n!}}\).
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. 0
Chọn C
Ta có: \(A = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {4x + 1} - 1}}{x} - \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{2x + 1}} - 1}}{x}\)
Mà: \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {4x + 1} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{4x}}{{x\left( {\sqrt {4x + 1} + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{4}{{\sqrt {4x + 1} + 1}} = 2\)
\(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{2x + 1}} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{x\left[ {\sqrt[3]{{{{(2x + 1)}^2}}} + \sqrt[3]{{2x + 1}} + 1} \right]}} = \frac{2}{3}\)
Vậy \(A = 2 - \frac{2}{3} = \frac{4}{3}\).
Ta có: \(A = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {4x + 1} - 1}}{x} - \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{2x + 1}} - 1}}{x}\)
Mà: \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {4x + 1} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{4x}}{{x\left( {\sqrt {4x + 1} + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{4}{{\sqrt {4x + 1} + 1}} = 2\)
\(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{2x + 1}} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{x\left[ {\sqrt[3]{{{{(2x + 1)}^2}}} + \sqrt[3]{{2x + 1}} + 1} \right]}} = \frac{2}{3}\)
Vậy \(A = 2 - \frac{2}{3} = \frac{4}{3}\).
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. \(\frac{2}{5}\)
Chọn D
Ta có: \(B = \mathop {\lim }\limits_{x \to 1} \frac{{4(x - 1)\left[ {\sqrt[3]{{{{(5x + 3)}^2}}} + 2\sqrt[3]{{5x + 3}} + 4} \right]}}{{5(x - 1)\left[ {\sqrt {4x + 5} + 3} \right]}}\)\( = \mathop {\lim }\limits_{x \to 1} \frac{{4\left[ {\sqrt[3]{{{{(5x + 3)}^2}}} + 2\sqrt[3]{{5x + 3}} + 4} \right]}}{{5\left( {\sqrt {4x + 5} + 3} \right)}} = \frac{2}{5}\).
Ta có: \(B = \mathop {\lim }\limits_{x \to 1} \frac{{4(x - 1)\left[ {\sqrt[3]{{{{(5x + 3)}^2}}} + 2\sqrt[3]{{5x + 3}} + 4} \right]}}{{5(x - 1)\left[ {\sqrt {4x + 5} + 3} \right]}}\)\( = \mathop {\lim }\limits_{x \to 1} \frac{{4\left[ {\sqrt[3]{{{{(5x + 3)}^2}}} + 2\sqrt[3]{{5x + 3}} + 4} \right]}}{{5\left( {\sqrt {4x + 5} + 3} \right)}} = \frac{2}{5}\).
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. 3
Chọn D
Ta có: \(C = \mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt[4]{{2x + 3}} - 1}}{{\sqrt {x + 2} - 1}} - \mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt[3]{{3x + 2}} + 1}}{{\sqrt {x + 2} - 1}}\)
\( = \mathop {\lim }\limits_{x \to - 1} \frac{{\frac{{\sqrt[4]{{2(x + 1) + 1}} - 1}}{{x + 1}}}}{{\frac{{\sqrt {(x + 1) + 1} - 1}}{{x + 1}}}} - \mathop {\lim }\limits_{x \to - 1} \frac{{\frac{{\sqrt[3]{{ - 3(x + 1) + 1}} - 1}}{{x + 1}}}}{{\frac{{\sqrt {(x + 1) + 1} - 1}}{{x + 1}}}} = \frac{{\frac{2}{4}}}{{\frac{1}{2}}} - \frac{{ - 1}}{{\frac{1}{2}}} = 3\)
Ta có: \(C = \mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt[4]{{2x + 3}} - 1}}{{\sqrt {x + 2} - 1}} - \mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt[3]{{3x + 2}} + 1}}{{\sqrt {x + 2} - 1}}\)
\( = \mathop {\lim }\limits_{x \to - 1} \frac{{\frac{{\sqrt[4]{{2(x + 1) + 1}} - 1}}{{x + 1}}}}{{\frac{{\sqrt {(x + 1) + 1} - 1}}{{x + 1}}}} - \mathop {\lim }\limits_{x \to - 1} \frac{{\frac{{\sqrt[3]{{ - 3(x + 1) + 1}} - 1}}{{x + 1}}}}{{\frac{{\sqrt {(x + 1) + 1} - 1}}{{x + 1}}}} = \frac{{\frac{2}{4}}}{{\frac{1}{2}}} - \frac{{ - 1}}{{\frac{1}{2}}} = 3\)
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. 1
Chọn D
Ta có: \(D = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {{x^2} - x - 2} \right)\left[ {{x^2} + x.\sqrt[3]{{3x + 2}} + \sqrt[3]{{{{(3x + 2)}^2}}}} \right]}}{{({x^3} - 3x - 2)\left( {x + \sqrt {x + 2} } \right)}}\)\( = \mathop {\lim }\limits_{x \to 2} \frac{{\left[ {{x^2} + x.\sqrt[3]{{3x + 2}} + \sqrt[3]{{{{(3x + 2)}^2}}}} \right]}}{{(x + 1)\left( {x + \sqrt {x + 2} } \right)}} = 1\).
Ta có: \(D = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {{x^2} - x - 2} \right)\left[ {{x^2} + x.\sqrt[3]{{3x + 2}} + \sqrt[3]{{{{(3x + 2)}^2}}}} \right]}}{{({x^3} - 3x - 2)\left( {x + \sqrt {x + 2} } \right)}}\)\( = \mathop {\lim }\limits_{x \to 2} \frac{{\left[ {{x^2} + x.\sqrt[3]{{3x + 2}} + \sqrt[3]{{{{(3x + 2)}^2}}}} \right]}}{{(x + 1)\left( {x + \sqrt {x + 2} } \right)}} = 1\).
A. + ∞
B. - ∞
C. 1/2
D. 0
Chọn C
Cách 1: Đặt \(t = \sqrt[3]{{3x + 1}} \Rightarrow x = \frac{{{t^3} - 1}}{3}\) và \(x \to 0 \Leftrightarrow t \to 1\)
Nên \(A = \mathop {\lim }\limits_{t \to 1} \frac{{\sqrt {1 + \frac{{{t^3} - 1}}{3}} - t}}{{{{\left( {\frac{{{t^3} - 1}}{3}} \right)}^2}}} = 9\mathop {\lim }\limits_{t \to 1} \frac{{\sqrt {\frac{{{t^3} + 2}}{3}} - t}}{{{{(t - 1)}^2}{{({t^2} + t + 1)}^2}}}\)
\( = 3\mathop {\lim }\limits_{t \to 1} \frac{{{t^3} - 3{t^2} + 2}}{{{{(t - 1)}^2}{{({t^2} + t + 1)}^2}\left( {\sqrt {\frac{{{t^3} + 2}}{3}} + t} \right)}}\)
\( = 3\mathop {\lim }\limits_{t \to 1} \frac{{{{(t - 1)}^2}(t + 2)}}{{{{(t - 1)}^2}{{({t^2} + t + 1)}^2}\left( {\sqrt {\frac{{{t^3} + 2}}{3}} + t} \right)}}\)
\( = 3\mathop {\lim }\limits_{t \to 1} \frac{{t + 2}}{{{{({t^2} + t + 1)}^2}\left( {\sqrt {\frac{{{t^3} + 2}}{3}} + t} \right)}} = \frac{1}{2}\).
Cách 2: Ta có:
\(A = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2x} - (1 + x)}}{{{x^2}}} - \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{1 + 3x}} - (1 + x)}}{{{x^2}}}\)
\( = \mathop {\lim }\limits_{x \to 0} \frac{{ - 1}}{{\sqrt {1 + 2x} + 1 + x}} - \mathop {\lim }\limits_{x \to 0} \frac{{ - 3 - x}}{{\sqrt[3]{{{{(1 + 3x)}^2}}} + (1 + x)\sqrt[3]{{1 + 3x}} + {{(1 + x)}^2}}}\)
Do đó: \(A = \frac{1}{2}\).
Cách 1: Đặt \(t = \sqrt[3]{{3x + 1}} \Rightarrow x = \frac{{{t^3} - 1}}{3}\) và \(x \to 0 \Leftrightarrow t \to 1\)
Nên \(A = \mathop {\lim }\limits_{t \to 1} \frac{{\sqrt {1 + \frac{{{t^3} - 1}}{3}} - t}}{{{{\left( {\frac{{{t^3} - 1}}{3}} \right)}^2}}} = 9\mathop {\lim }\limits_{t \to 1} \frac{{\sqrt {\frac{{{t^3} + 2}}{3}} - t}}{{{{(t - 1)}^2}{{({t^2} + t + 1)}^2}}}\)
\( = 3\mathop {\lim }\limits_{t \to 1} \frac{{{t^3} - 3{t^2} + 2}}{{{{(t - 1)}^2}{{({t^2} + t + 1)}^2}\left( {\sqrt {\frac{{{t^3} + 2}}{3}} + t} \right)}}\)
\( = 3\mathop {\lim }\limits_{t \to 1} \frac{{{{(t - 1)}^2}(t + 2)}}{{{{(t - 1)}^2}{{({t^2} + t + 1)}^2}\left( {\sqrt {\frac{{{t^3} + 2}}{3}} + t} \right)}}\)
\( = 3\mathop {\lim }\limits_{t \to 1} \frac{{t + 2}}{{{{({t^2} + t + 1)}^2}\left( {\sqrt {\frac{{{t^3} + 2}}{3}} + t} \right)}} = \frac{1}{2}\).
Cách 2: Ta có:
\(A = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2x} - (1 + x)}}{{{x^2}}} - \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{1 + 3x}} - (1 + x)}}{{{x^2}}}\)
\( = \mathop {\lim }\limits_{x \to 0} \frac{{ - 1}}{{\sqrt {1 + 2x} + 1 + x}} - \mathop {\lim }\limits_{x \to 0} \frac{{ - 3 - x}}{{\sqrt[3]{{{{(1 + 3x)}^2}}} + (1 + x)\sqrt[3]{{1 + 3x}} + {{(1 + x)}^2}}}\)
Do đó: \(A = \frac{1}{2}\).
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. - 1
Chọn D
Ta có: \(B = \mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt {5 + 4x} - \sqrt[3]{{7 + 6x}}}}{{{{\left( {x + 1} \right)}^2}\left( {x - 1} \right)}}\)
Đặt \(t = x + 1\). Khi đó:
\(\mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt {5 + 4x} - \sqrt[3]{{7 + 6x}}}}{{{{\left( {x + 1} \right)}^2}}} = \mathop {\lim }\limits_{t \to 0} \frac{{\sqrt {1 + 4t} - \sqrt[3]{{1 + 6t}}}}{{{t^2}}}\)
\( = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 4t} - (2t + 1)}}{{{t^2}}} - \mathop {\lim }\limits_{t \to 0} \frac{{\sqrt[3]{{1 + 6t}} - (2t + 1)}}{{{t^2}}}\)
\( = \mathop {\lim }\limits_{t \to 0} \frac{{ - 4}}{{\sqrt {1 + 4t} + 2t + 1}} - \mathop {\lim }\limits_{t \to 0} \frac{{ - 8t - 12}}{{\sqrt[3]{{{{(1 + 6t)}^2}}} + (2t + 1)\sqrt[3]{{{{(1 + 6t)}^2}}} + {{(2t + 1)}^2}}}\)\( = 2\).
Do đó: \(B = - 1\).
Ta có: \(B = \mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt {5 + 4x} - \sqrt[3]{{7 + 6x}}}}{{{{\left( {x + 1} \right)}^2}\left( {x - 1} \right)}}\)
Đặt \(t = x + 1\). Khi đó:
\(\mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt {5 + 4x} - \sqrt[3]{{7 + 6x}}}}{{{{\left( {x + 1} \right)}^2}}} = \mathop {\lim }\limits_{t \to 0} \frac{{\sqrt {1 + 4t} - \sqrt[3]{{1 + 6t}}}}{{{t^2}}}\)
\( = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 4t} - (2t + 1)}}{{{t^2}}} - \mathop {\lim }\limits_{t \to 0} \frac{{\sqrt[3]{{1 + 6t}} - (2t + 1)}}{{{t^2}}}\)
\( = \mathop {\lim }\limits_{t \to 0} \frac{{ - 4}}{{\sqrt {1 + 4t} + 2t + 1}} - \mathop {\lim }\limits_{t \to 0} \frac{{ - 8t - 12}}{{\sqrt[3]{{{{(1 + 6t)}^2}}} + (2t + 1)\sqrt[3]{{{{(1 + 6t)}^2}}} + {{(2t + 1)}^2}}}\)\( = 2\).
Do đó: \(B = - 1\).
Sửa lần cuối: