Chủ đề này có 25 trả lời

[tolaphuy10a1lhp]

1) $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{3^n} - {4^n} + {5^n}}}{{{3^n} + {4^n} - {5^n}}}$

2) $\lim \dfrac{{1 + a + {a^2} + ... + {a^n}}}{{1 + b + {b^2} + ... + {b^n}}}\left( {\left| a \right| < 1;\left| b \right| < 1} \right)$

3)$\lim \left( {\dfrac{1}{{\sqrt {{n^2} + 1} }} + \dfrac{1}{{\sqrt {{n^2} + 2} }} + ... + \dfrac{1}{{\sqrt {{n^2} + n} }}} \right)$

Bài viết đã được chỉnh sửa nội dung bởi tolaphuy10a1lhp: 24-12-2011 - 15:07

[Crystal]

1) $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{3^n} - {4^n} + {5^n}}}{{{3^n} + {4^n} - {5^n}}}$

Giải:

$$1)\,\,\mathop {\lim }\limits_{n \to \infty } \dfrac{{{3^n} - {4^n} + {5^n}}}{{{3^n} + {4^n} - {5^n}}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{{{\left( {\dfrac{3}{5}} \right)}^n} - {{\left( {\dfrac{4}{5}} \right)}^n} + 1}}{{{{\left( {\dfrac{3}{5}} \right)}^n} + {{\left( {\dfrac{4}{5}} \right)}^n} - 1}} = \boxed{ - 1}$$
Do $0 < \dfrac{3}{5},\dfrac{4}{5} < 1$ nên ${\left( {\dfrac{3}{5}} \right)^n} \to 0,{\left( {\dfrac{4}{5}} \right)^n} \to 0\,\,\,khi\,\,n \to \infty $

[E. Galois]

2) $$\lim \dfrac{{1 + a + {a^2} + ... + {a^n}}}{{1 + b + {b^2} + ... + {b^n}}}=\lim\left (\dfrac{1-a^n}{1-a}. \dfrac{1-b}{1-b^n} \right )=\dfrac{1-b}{1-a}$$
Vì $\lim a^n=\lim b^n = 0$ do $\left( {\left| a \right| < 1;\left| b \right| < 1} \right)$

[Crystal]

3)$\lim \left( {\dfrac{1}{{\sqrt {{n^2} + 1} }} + \dfrac{1}{{\sqrt {{n^2} + 2} }} + ... + \dfrac{1}{{\sqrt {{n^2} + n} }}} \right)$

Đặt $$L = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{1}{{\sqrt {{n^2} + 1} }} + \dfrac{1}{{\sqrt {{n^2} + 2} }} + ... + \dfrac{1}{{\sqrt {{n^2} + n} }}} \right) = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\dfrac{1}{{\sqrt {{n^2} + i} }}} $$
Ta có: $$\dfrac{n}{{\sqrt {{n^2} + n} }} \leqslant \sum\limits_{i = 1}^n {\dfrac{1}{{\sqrt {{n^2} + i} }}} \leqslant \dfrac{n}{{\sqrt {{n^2} + 1} }}\,,\,\,\,\left( {1 \leqslant i \leqslant n} \right)$$
Mặt khác: $$\mathop {\lim }\limits_{n \to \infty } \dfrac{n}{{\sqrt {{n^2} + n} }} = \mathop {\lim }\limits_{n \to \infty } \dfrac{n}{{n\sqrt {1 + \dfrac{1}{n}} }} = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{\sqrt {1 + \dfrac{1}{n}} }} = 1$$
$$\mathop {\lim }\limits_{n \to \infty } \dfrac{n}{{\sqrt {{n^2} + 1} }} = \mathop {\lim }\limits_{n \to \infty } \dfrac{n}{{n\sqrt {1 + \dfrac{1}{{{n^2}}}} }} = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{\sqrt {1 + \dfrac{1}{{{n^2}}}} }} = 1$$
Theo nguyên lí kẹp, suy ra: $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\dfrac{1}{{\sqrt {{n^2} + i} }}} = 1$$
Vậy $\boxed{L = 1}$

[tolaphuy10a1lhp]

Thêm một số bài, mới học nên chưa quen !

4) $\lim \left( {\sqrt {{n^2} + 2n} - n + 1} \right)$

5) $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 - 2x} - \sqrt[3]{{1 - 3x}}}}{{{x^2}}}$

6) $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {2x + 1} - \sqrt[3]{{2x + 1}}}}{{\sqrt[3]{{2x + 1}} - 1}}$

7) $\mathop {\lim }\limits_{x \to 1} \dfrac{{x + {x^2} + {x^3} + ... + {x^{2011}} - 2011}}{{x - 1}}$

[Crystal]

Thêm một số bài, mới học nên chưa quen !

4) $\lim \left( {\sqrt {{n^2} + 2n} - n + 1} \right)$

5) $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 - 2x} - \sqrt[3]{{1 - 3x}}}}{{{x^2}}}$

4. $$\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt {{n^2} + 2n} - n + 1} \right) = \mathop {\lim }\limits_{n \to \infty } \dfrac{{{n^2} + 2n - {{\left( {n - 1} \right)}^2}}}{{\sqrt {{n^2} + 2n} + n - 1}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{4n - 1}}{{\sqrt {{n^2} + 2n} + n - 1}}$$
$$ = \mathop {\lim }\limits_{n \to \infty } \dfrac{{n\left( {4 - \dfrac{1}{n}} \right)}}{{n\left( {\sqrt {1 + \dfrac{2}{n}} + 1 - \dfrac{1}{n}} \right)}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{4 - \dfrac{1}{n}}}{{\sqrt {1 + \dfrac{2}{n}} + 1 - \dfrac{1}{n}}} = \boxed2$$

5) Ta có: $$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 - 2x} - \sqrt[3]{{1 - 3x}}}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{1}{x}\left( {\dfrac{{\sqrt {1 - 2x} - 1}}{x} - \dfrac{{\sqrt[3]{{1 - 3x}} - 1}}{x}} \right)} \right]$$
$$ = \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{1}{x}\left( {\dfrac{{ - 2}}{{\sqrt {1 - 2x} + 1}} + \dfrac{3}{{\sqrt[3]{{{{\left( {1 - 3x} \right)}^2}}} + \sqrt[3]{{1 - 3x}} + 1}}} \right)} \right]$$
$$ = \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{1}{x}\left( {\dfrac{{3\left( {\sqrt {1 - 2x} + 1} \right) - 2\left( {\sqrt[3]{{{{\left( {1 - 3x} \right)}^2}}} + \sqrt[3]{{1 - 3x}} + 1} \right)}}{{\left( {\sqrt {1 - 2x} + 1} \right)\left( {\sqrt[3]{{{{\left( {1 - 3x} \right)}^2}}} + \sqrt[3]{{1 - 3x}} + 1} \right)}}} \right)} \right]$$
$$ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{3\dfrac{{\sqrt {1 - 2x} - 1}}{x} - 2\dfrac{{\sqrt[3]{{{{\left( {1 - 3x} \right)}^2}}} - 1}}{x} - 2\dfrac{{\sqrt[3]{{1 - 3x}} - 1}}{x}}}{{\left( {\sqrt {1 - 2x} + 1} \right)\left( {\sqrt[3]{{{{\left( {1 - 3x} \right)}^2}}} + \sqrt[3]{{1 - 3x}} + 1} \right)}}} \right)$$
Lại có: $$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 - 2x} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 2}}{{\sqrt {1 - 2x} + 1}} = \boxed{ - 1}$$
$$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt[3]{{{{\left( {1 - 3x} \right)}^2}}} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt[3]{{1 - 3x}} + 1} \right)\dfrac{{\sqrt[3]{{1 - 3x}} - 1}}{x} = 2\mathop {\lim }\limits_{x \to 0} \dfrac{{ - 3}}{{\sqrt[3]{{{{\left( {1 - 3x} \right)}^2}}} + \sqrt[3]{{1 - 3x}} + 1}} = \boxed{ - 2}$$
$$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt[3]{{1 - 3x}} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 3}}{{\sqrt[3]{{{{\left( {1 - 3x} \right)}^2}}} + \sqrt[3]{{1 - 3x}} + 1}} = \boxed{ - 1}$$
Vậy $$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 - 2x} - \sqrt[3]{{1 - 3x}}}}{{{x^2}}} = \dfrac{{3\left( { - 1} \right) - 2\left( { - 2} \right) - 2\left( { - 1} \right)}}{6} = \boxed{\dfrac{1}{2}}$$

Bài viết đã được chỉnh sửa nội dung bởi xusinst: 21-12-2011 - 22:20

[Crystal]

6) $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {2x + 1} - \sqrt[3]{{2x + 1}}}}{{\sqrt[3]{{2x + 1}} - 1}}$

7) $\mathop {\lim }\limits_{x \to 1} \dfrac{{x + {x^2} + {x^3} + ... + {x^{2011}} - 2011}}{{x - 1}}$

Bài 6: Cách làm tương tự bài 5.

Bài 7:

$$\lim_{x\rightarrow 1}\dfrac{x+x^{2}+...+x^{n}-n}{x-1}$$

Ta có: $$x + {x^2} + ... + {x^n} - n = \left( {x - 1} \right) + \left( {{x^2} - 1} \right) + ... + \left( {{x^n} - 1} \right)$$
$$ = \left( {x - 1} \right)\left( {1 + \left( {x + 1} \right) + \left( {{x^2} + x + 1} \right) + ... + \left( {{x^{n - 1}} + {x^{n - 2}} + ... + x + 1} \right)} \right)$$
Do đó: $$\mathop {\lim }\limits_{x \to 1} \dfrac{{x + {x^2} + ... + {x^n} - n}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left( {1 + \left( {x + 1} \right) + \left( {{x^2} + x + 1} \right) + ... + \left( {{x^{n - 1}} + {x^{n - 2}} + ... + x + 1} \right)} \right)$$
$$ = 1 + 2 + 3 + ... + n = \boxed{\dfrac{{n\left( {n + 1} \right)}}{2}}$$

Cho $n=2011$, ta được: $$\mathop {\lim }\limits_{x \to 1} \dfrac{{x + {x^2} + {x^3} + ... + {x^{2011}} - 2011}}{{x - 1}} = \boxed{2023066}$$

[tolaphuy10a1lhp]

Thêm vài bài:
8) $\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {x + 3} .\sqrt[3]{{x + 7}} - 4}}{{{x^2} - 1}}$

9) $\lim \dfrac{1}{{\sqrt n }}\left( {1 + \dfrac{1}{{\sqrt 2 }} + ... + \dfrac{1}{{\sqrt n }}} \right)$

10) $\mathop {\lim }\limits_{x \to a} \dfrac{{\left( {{x^{2011}} - {a^{2011}}} \right) - 2011{a^{2010}}\left( {x - a} \right)}}{{{{\left( {x - a} \right)}^2}}}$

11) $\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos x\cos 2x\cos 3x...\cos 10x}}{{{x^2}}}$

Cảm ơn Anh Xusinst nhiều. Đúng là bài 10 em sai đề. Đã sửa lại.

Bài viết đã được chỉnh sửa nội dung bởi tolaphuy10a1lhp: 27-12-2011 - 22:15

[hathanh123]

Thêm vài bài:
8) $\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {x + 3} .\sqrt[3]{{x + 7}} - 4}}{{{x^2} - 1}}$

9) $\lim \dfrac{1}{{\sqrt n }}\left( {1 + \dfrac{1}{{\sqrt 2 }} + ... + \dfrac{1}{{\sqrt n }}} \right)$

10) $\mathop {\lim }\limits_{x \to a} \dfrac{{\left( {{x^{2011}} - {a^{2010}}} \right) - 2011{a^{2010}}\left( {x - a} \right)}}{{{{\left( {x - a} \right)}^2}}}$

11) $\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos x\cos 2x\cos 3x...\cos 10x}}{{{x^2}}}$

Mấy bài này sao rắc rối quá?

[Crystal]

Thêm vài bài:
9) $\lim \dfrac{1}{{\sqrt n }}\left( {1 + \dfrac{1}{{\sqrt 2 }} + ... + \dfrac{1}{{\sqrt n }}} \right)$

10) $\mathop {\lim }\limits_{x \to a} \dfrac{{\left( {{x^{2011}} - {a^{2010}}} \right) - 2011{a^{2010}}\left( {x - a} \right)}}{{{{\left( {x - a} \right)}^2}}}$

11) $\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos x\cos 2x\cos 3x...\cos 10x}}{{{x^2}}}$

Câu 9: Dùng định lí Stolz

Đặt $$\left\{ \begin{array}{l}
{x_n} = 1 + \dfrac{1}{{\sqrt 2 }} + ... + \dfrac{1}{{\sqrt n }}\\
{y_n} = \sqrt n
\end{array} \right.\,\,\,\,\,\left( {n \ge 1} \right)$$
Khi đó: dãy $\left\{ {{y_n}} \right\}$ tăng thực sự tới $ + \infty $ tức $\mathop {\lim }\limits_{n \to + \infty } {y_n} = + \infty $

Ta có: $$\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{x_n} - {x_{n - 1}}}}{{{y_n} - {y_{n - 1}}}} = \mathop {\lim }\limits_{n \to + \infty } \dfrac{{\dfrac{1}{{\sqrt n }}}}{{\sqrt n - \sqrt {n - 1} }} = \mathop {\lim }\limits_{n \to + \infty } \dfrac{{\sqrt n + \sqrt {n - 1} }}{{\sqrt n }} = 2$$
Theo định lí Stolz, suy ra: $$\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{x_n}}}{{{y_n}}} = \mathop {\lim }\limits_{n \to + \infty } \dfrac{1}{{\sqrt n }}\left( {1 + \dfrac{1}{{\sqrt 2 }} + ... + \dfrac{1}{{\sqrt n }}} \right) = \boxed2$$

Câu 10: Anh nghĩ đề đúng phải là $$\mathop {\lim }\limits_{x \to a} \dfrac{{\left( {{x^{2011}} - {a^{2011}}} \right) - 2011{a^{2010}}\left( {x - a} \right)}}{{{{\left( {x - a} \right)}^2}}}$$

Khi đó, anh đưa ra bài toán tổng quát: $$\boxed{\mathop {\lim }\limits_{x \to a} \dfrac{{\left( {{x^n} - {a^n}} \right) - n{a^{n - 1}}\left( {x - a} \right)}}{{{{\left( {x - a} \right)}^2}}}}$$

Ta có: $$\left( {{x^n} - {a^n}} \right) - n{a^{n - 1}}\left( {x - a} \right) = \left( {x - a} \right)\left( {{x^{n - 1}} + a{x^{n - 2}} + ... + {a^{n - 1}}} \right) - n{a^{n - 1}}\left( {x - a} \right)$$
$$ = \left( {x - a} \right)\left[ {\left( {{x^{n - 1}} + a{x^{n - 2}} + ... + {a^{n - 1}}} \right) - n{a^{n - 1}}} \right]$$
Lại có: $${x^{n - 1}} + a{x^{n - 2}} + ... + {a^{n - 1}} - n{a^{n - 1}} = \left( {{x^{n - 1}} - {a^{n - 1}}} \right) + \left( {a{x^{n - 2}} - {a^{n - 1}}} \right) + ... + \left( {x{a^{n - 2}} - {a^{n - 1}}} \right)$$
$$ = \left( {x - a} \right)\left( {{x^{n - 2}} + ... + {a^{n - 2}}} \right) + a\left( {{x^{n - 2}} - {a^{n - 2}}} \right) + ... + {a^{n - 2}}\left( {x - a} \right)$$
$$ = \left( {x - a} \right)\left[ {\left( {{x^{n - 2}} + ... + {a^{n - 2}}} \right) + a\left( {{x^{n - 3}} + ... + {a^{n - 3}}} \right) + ... + {a^{n - 2}}} \right]$$
Do đó: $$\left( {{x^n} - {a^n}} \right) - n{a^{n - 1}}\left( {x - a} \right) = {\left( {x - a} \right)^2}\left[ {\left( {{x^{n - 2}} + ... + {a^{n - 2}}} \right) + a\left( {{x^{n - 3}} + ... + {a^{n - 3}}} \right) + ... + {a^{n - 2}}} \right]$$
Vậy $$\mathop {\lim }\limits_{x \to a} \dfrac{{\left( {{x^n} - {a^n}} \right) - n{a^{n - 1}}\left( {x - a} \right)}}{{{{\left( {x - a} \right)}^2}}} = \mathop {\lim }\limits_{x \to a} \left[ {\left( {{x^{n - 2}} + ... + {a^{n - 2}}} \right) + a\left( {{x^{n - 3}} + ... + {a^{n - 3}}} \right) + ... + {a^{n - 2}}} \right]$$
$$ = \boxed{\dfrac{{n\left( {n - 1} \right)}}{2}{a^{n - 2}}}$$
Từ đó, cho $n = 2011$, ta được: $$\mathop {\lim }\limits_{x \to a} \dfrac{{\left( {{x^{2011}} - {a^{2011}}} \right) - 2011{a^{2010}}\left( {x - a} \right)}}{{{{\left( {x - a} \right)}^2}}} = \boxed{\dfrac{{2010.2011}}{2}{a^{2009}}}$$

[Crystal]

11) $\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos x\cos 2x\cos 3x...\cos 10x}}{{{x^2}}}$

Đặt $$A = 1 - \cos x\cos 2x\cos 3x...\cos 10x$$
Ta có: $$A = \left( {1 - \cos x} \right)\cos 2x\cos 3x...\cos 10 + \left( {1 - \cos 2x} \right)\cos 3x...\cos 10 + $$
$$ + \left( {1 - \cos 3x} \right)\cos 4x...\cos 10x + ... + \left( {1 - \cos 10x} \right)$$
Lại có: $$\dfrac{{1 - \cos kx}}{{{x^2}}} = \dfrac{{2{{\sin }^2}\left( {\dfrac{{kx}}{2}} \right)}}{{{x^2}}} = \dfrac{{{{\sin }^2}\left( {\dfrac{{kx}}{2}} \right)}}{{{{\left( {\dfrac{{kx}}{2}} \right)}^2}}}.\dfrac{{{k^2}}}{2}$$
Suy ra: $$\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos kx}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}\left( {\dfrac{{kx}}{2}} \right)}}{{{{\left( {\dfrac{{kx}}{2}} \right)}^2}}}.\dfrac{{{k^2}}}{2} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}\left( {\dfrac{{kx}}{2}} \right)}}{{{{\left( {\dfrac{{kx}}{2}} \right)}^2}}}.\mathop {\lim }\limits_{x \to 0} \dfrac{{{k^2}}}{2} = \boxed{\dfrac{{{k^2}}}{2}}$$
và $$\mathop {\lim }\limits_{x \to 0} \cos 2x = \mathop {\lim }\limits_{x \to 0} \cos 3x = ... = \mathop {\lim }\limits_{x \to 0} \cos 10x = \boxed1$$
Do đó: $$\mathop {\lim }\limits_{x \to 0} A = \dfrac{1}{2} + \dfrac{{{2^2}}}{2} + \dfrac{{{3^2}}}{2} + ... + \dfrac{{{{10}^2}}}{2} = \dfrac{1}{2}\sum\limits_{k = 1}^{10} {{k^2}} = \dfrac{1}{2}\left( {\dfrac{{10.11.21}}{6}} \right) = \boxed{\dfrac{{385}}{2}}$$

[Crystal]

8) $\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {x + 3} .\sqrt[3]{{x + 7}} - 4}}{{{x^2} - 1}}$

Giải quyết nốt câu này.

Ta có: $$\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {x + 3} \sqrt[3]{{x + 7}} - 4}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {\sqrt {x + 3} - 2} \right)\sqrt[3]{{x + 7}} + 2\left( {\sqrt[3]{{x + 7}} - 2} \right)}}{{{x^2} - 1}}$$
$$ = \mathop {\lim }\limits_{x \to 1} \left( {\dfrac{{\left( {\sqrt {x + 3} - 2} \right)\sqrt[3]{{x + 7}}}}{{{x^2} - 1}} + 2\dfrac{{\sqrt[3]{{x + 7}} - 2}}{{{x^2} - 1}}} \right) = 2\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {x + 3} - 2}}{{{x^2} - 1}} + 2\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt[3]{{x + 7}} - 2}}{{{x^2} - 1}}$$
Tính: $$\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {x + 3} - 2}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{x - 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {\sqrt {x + 3} + 2} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{1}{{\left( {x + 1} \right)\left( {\sqrt {x + 3} + 2} \right)}} = \boxed{\dfrac{1}{8}}$$
Tính: $$\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt[3]{{x + 7}} - 2}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{x - 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {\sqrt[3]{{{{\left( {x + 7} \right)}^2}}} + 2\sqrt[3]{{x + 7}} + 4} \right)}}$$
$$ = \mathop {\lim }\limits_{x \to 1} \dfrac{1}{{\left( {x + 1} \right)\left( {\sqrt[3]{{{{\left( {x + 7} \right)}^2}}} + 2\sqrt[3]{{x + 7}} + 4} \right)}} = \boxed{\dfrac{1}{{24}}}$$
Do đó: $$\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {x + 3} \sqrt[3]{{x + 7}} - 4}}{{{x^2} - 1}} = \dfrac{2}{8} + \dfrac{2}{{24}} = \boxed{\dfrac{1}{3}}$$

[tocxu]

anh xusint giúp em bài này nhá
Tìm $L = \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}}}{{\sqrt {1 + x\sin 3x} - \sqrt {\cos 2x} }}$

[Crystal]

anh xusint giúp em bài này nhá
Tìm $L = \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}}}{{\sqrt {1 + x\sin 3x} - \sqrt {\cos 2x} }}$

@ tocxu: Lần sau em viết đúng tên anh nhé. Anh xin ra tay luôn với bài này

Ta có: $$\sqrt {1 + x\sin 3x} - \sqrt {\cos 2x} = \left( {\sqrt {1 + x\sin 3x} - 1} \right) + \left( {1 - \sqrt {\cos 2x} } \right)$$
$$ = \dfrac{{x\sin 3x}}{{\sqrt {1 + x\sin 3x} + 1}} + \dfrac{{1 - \cos 2x}}{{1 + \sqrt {\cos 2x} }} = \dfrac{{x\sin 3x}}{{\sqrt {1 + x\sin 3x} + 1}} + \dfrac{{2{{\sin }^2}x}}{{1 + \sqrt {\cos 2x} }}$$
Suy ra: $$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + x\sin 3x} - \sqrt {\cos 2x} }}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{x\sin 3x}}{{{x^2}\left( {\sqrt {1 + x\sin 3x} + 1} \right)}} + \dfrac{{2{{\sin }^2}x}}{{{x^2}\left( {1 + \sqrt {\cos 2x} } \right)}}} \right)$$
$$ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin 3x}}{{3x}}.\dfrac{3}{{\sqrt {1 + x\sin 3x} + 1}}} \right) + \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{{\sin }^2}x}}{{{x^2}}}.\dfrac{2}{{1 + \sqrt {\cos 2x} }}} \right) = \dfrac{3}{2} + 1 = \boxed{\dfrac{5}{2}}$$
Vậy $$\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}}}{{\sqrt {1 + x\sin 3x} - \sqrt {\cos 2x} }} = \boxed{\dfrac{2}{5}}$$

[tocxu]

cảm ơn anh xusinst nhiều, lần sau em sẽ chú ý

[Crystal]

Anh đưa ra bài toán tổng quát cho bài 11.

Bài toán tổng quát [11]: Tìm $$\lim_{x\rightarrow 0}\dfrac{1-cosxcos2xcos3x...cosnx}{x^{2}},\; \forall n\in \mathbb{N}^{*}$$

Giải:

Đặt $${a_n} =\lim_{x\rightarrow 0}\dfrac{1-cosxcos2xcos3x...cosnx}{x^{2}}$$
Suy ra: $$a_{n+1}=\lim_{x\rightarrow 0}\dfrac{1-cosxcos2xcos3x...cos\left (n+1 \right )x}{x^{2}}$$
$$=\lim_{x\rightarrow 0}\dfrac{1-cosxcos2xcos3x...cosnx}{x^{2}}+\lim_{x\rightarrow 0}\dfrac{cosxcos2xcos3x...cosnx\left ( 1-cos\left ( n+1 \right )x \right )}{x^{2}}$$
$$={a_n}+\lim_{x\rightarrow 0}\dfrac{1-cos\left ( n+1 \right )x}{x^{2}}={a_n}+\lim_{x\rightarrow 0}\dfrac{2sin^{2}\dfrac{\left ( n+1 \right )x}{2}}{x^{2}}$$
$$={a_n}+\dfrac{\left ( n+1 \right )^{2}}{2}\lim_{x\rightarrow 0}\left ( \dfrac{sin\dfrac{\left ( n+1 \right )x}{2}}{\dfrac{n+1}{2}} \right )^{2}={a_n}+\dfrac{\left ( n+1 \right )^{2}}{2}$$
Cho $n$ nhận các giá trị từ $1$ đến $n-1$, ta có:
$$a_{0}=0$$
$$a_{1}=a_{0}+\dfrac{1}{2}$$
$$a_{2}=a_{1}+\dfrac{2^{2}}{2}$$
$$a_{3}=a_{2}+\dfrac{3^{2}}{2}$$
$$........$$
$${a_n}=a_{n-1}+\dfrac{n^{2}}{2}$$
Từ đó ta nhận được: $${a_n}=\sum_{k=1}^{n}\left ( \dfrac{k^{2}}{2} \right )=\dfrac{1}{2}\left ( \sum_{k=1}^{n}k^{2} \right )=\dfrac{1}{2}\dfrac{n\left ( n+1 \right )\left ( 2n+1 \right )}{6}$$
Vậy $$\lim_{x\rightarrow 0}\dfrac{1-cosxcos2xcos3x...cosnx}{x^{2}}=\boxed{\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{12}}}$$

Bài viết đã được chỉnh sửa nội dung bởi xusinst: 30-12-2011 - 17:32

[tolaphuy10a1lhp]

Chúc mừng Anh Xusinst .
Tiếp vài bài:
13) $\lim \left[ {\dfrac{{1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + ... + \dfrac{1}{{{2^n}}}}}{{1 + \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + ... + \dfrac{1}{{{5^n}}}}}} \right]$
14) $\lim \left[ {\dfrac{1}{{1.2.3.4}} + \dfrac{1}{{2.3.4.5}} + ... + \dfrac{1}{{n.\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}} \right]$
15) $\mathop {\lim }\limits_{n \to + \infty } \dfrac{1}{{1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n}}}$

[Crystal]

Chúc mừng Anh Xusinst .
13) $\lim \left[ {\dfrac{{1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + ... + \dfrac{1}{{{2^n}}}}}{{1 + \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + ... + \dfrac{1}{{{5^n}}}}}} \right]$

Cảm ơn em. Anh làm trước bài này.

Đặt $\left\{ \begin{array}{l}
A = 1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + ... + \dfrac{1}{{{2^n}}}\\
B = 1 + \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + ... + \dfrac{1}{{{5^n}}}
\end{array} \right.$

Nhận thấy $A$ là tổng của $n+1$ số hạng đầu tiên của cấp số nhân với số hạng đầu ${u_1} = 1$, công bội $q = \dfrac{1}{2}$, do đó:
$$A = {u_1}\dfrac{{1 - {q^{n + 1}}}}{{1 - q}} = \dfrac{{1 - {{\left( {\dfrac{1}{2}} \right)}^{n + 1}}}}{{1 - \dfrac{1}{2}}} = 2\left( {1 - {{\left( {\dfrac{1}{2}} \right)}^{n + 1}}} \right)$$
Tương tự, ta có: $B = 5\left( {1 - {{\left( {\dfrac{1}{5}} \right)}^{n + 1}}} \right)$
Khi đó: $$\mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + ... + \dfrac{1}{{{2^n}}}}}{{1 + \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + ... + \dfrac{1}{{{5^n}}}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \dfrac{A}{B} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{2\left( {1 - {{\left( {\dfrac{1}{2}} \right)}^{n + 1}}} \right)}}{{5\left( {1 - {{\left( {\dfrac{1}{5}} \right)}^{n + 1}}} \right)}} = \boxed{\dfrac{2}{5}}$$

[Crystal]

15) $\mathop {\lim }\limits_{n \to + \infty } \dfrac{1}{{1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n}}}$

Tiếp bài 15

Ta có đánh giá sau: $$\boxed{\ln \left( {1 + x} \right) < x,\,\,\forall x > 0}$$
Suy ra: $$\ln \left( {1 + \dfrac{1}{n}} \right) < \dfrac{1}{n},\,\forall n\in \mathbb{N^{*}} \Rightarrow \dfrac{1}{n} > \ln \left( {1 + n} \right) - \ln $$
hay $$1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n} > \left( {\ln 2 - \ln 1} \right) + \left( {\ln 3 - \ln 2} \right) + ... + \left[ {\ln \left( {n + 1} \right) - \ln n} \right] \geqslant \ln \left( {n + 1} \right)$$
Do đó: $$1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n} > \ln \left( {n + 1} \right),\forall n\in \mathbb{N^{*}}$$
Vậy $$\mathop {\lim }\limits_{n \to + \infty } \dfrac{1}{{1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n}}} = \boxed0$$
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Hoặc ta dùng đánh giá sau: $$\boxed{\mathop {\lim }\limits_{n \to + \infty } \left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n}} \right) = + \infty }$$
Từ đó dễ suy ra $$\mathop {\lim }\limits_{n \to + \infty } \dfrac{1}{{1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n}}} = \boxed0$$

[Crystal]

Anh tặng em bài này.

Bài 16: Tìm $$\boxed{\mathop {\lim }\limits_{n \to \infty } \sin \left( {n\pi \sqrt[3]{{{n^3} + 3{n^2} + 4n - 5}}} \right)}$$