Chương 4: GIỚI HẠN - Hỏi đáp

Lời giải:
a)

\(\lim\limits_{x\to-1}\frac{\sqrt[3]{x}+1}{2x^2+5x+3}=\lim\limits_{x\to-1}\frac{x+1}{\left(\sqrt[3]{x^2}-\sqrt[3]{x}+1\right)\left(x+1\right)\left(2x+3\right)}\)

\(\lim\limits_{x\to-1}\frac{1}{\left(\sqrt[3]{x^2}-\sqrt[3]{x}+1\right)\left(2x+3\right)}=\frac{1}{\left(\sqrt[3]{\left(-1\right)^2}-\sqrt[3]{-1}+1\right)\left(2.-1+3\right)}=\frac{1}{3}\)

b)

\(\lim\limits_{x\to1}\frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(\sqrt[3]{x}-1\right)^2}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(x-1\right)^2}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2\left(x-1\right)^2}\)

\(=\lim\limits_{x\to1}\frac{1}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2}=\frac{1}{\left(1+1+1\right)^2}=\frac{1}{9}\)

c)

\(\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{x^3+x^2-2}=\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{(x-1)(x^2+2x+2)}=\lim_{x\to 1}\frac{x-1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x-1)(x^2+2x+2)}\)

\(=\lim_{x\to 1}\frac{1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x^2+2x+2)}=\frac{1}{(1+1)(1+1)(1+2.1+2)}=\frac{1}{20}\)

d)

\(\lim_{x\to -2}\frac{\sqrt[3]{2x+12}+x}{x^2+2x}=\lim_{x\to -2}\frac{2x+12+x^3}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}\)

\(=\lim_{x\to -2}\frac{(x+2)(x^2-2x+6)}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}=\lim_{x\to -2}\frac{x^2-2x+6}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x}\)

\(=\frac{-7}{12}\)

Câu 3.

\(\lim \dfrac{{1 + a + {a^2} + ... + {a^n}}}{{1 + b + {b^2} + ... + {b^n}}} = \lim \dfrac{{\left( {1 + a + {a^2} + ... + {a^n}} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}{{\left( {1 + b + {b^2} + ... + {b^n}} \right)\left( {1 - b} \right)\left( {1 - a} \right)}} = \lim \dfrac{{\left( {1 - {a^{n + 1}}} \right)\left( {1 - b} \right)}}{{\left( {1 - {b^{n + 1}}} \right)\left( {1 - a} \right)}} = \dfrac{{1 - b}}{{1 - a}}\)

Câu 2.

\(\lim \dfrac{{3\sin n + 4\cos n}}{{n + 1}}\)

Vì \( - 1 \le \sin n \le 1; - 1 \le \cos n \le 1 \Rightarrow \) khi \(x \to \infty \) thì \(3\sin n + 4{\mathop{\rm cosn}\nolimits} = const \)

\(\Rightarrow T = \lim \dfrac{{3\sin n + 4\cos n}}{{n + 1}} = 0 \)

Chú thích: $const$ là kí hiệu hằng số, giống như dạng giới hạn L/vô cùng.

Câu 3.

Ta có biến đổi:

\(\lim \left( {\dfrac{{{n^2} - n}}{{1 - 2{n^2}}} + \dfrac{{2\sin {n^2}}}{{\sqrt n }}} \right) = \lim \dfrac{{{n^2} - n}}{{1 - 2{n^2}}} = \dfrac{1}{2}\)

Câu 1.

Vì \(\sqrt{2},\left(\sqrt{2}\right)^2,...,\left(\sqrt{2}\right)^n\) lập thành cấp số nhân có \(u_1=\sqrt{2}=q\) nên

\({u_n} = \sqrt 2 .\dfrac{{1 - {{\left( {\sqrt 2 } \right)}^n}}}{{1 - \sqrt 2 }} = \left( {2 - \sqrt 2 } \right)\left[ {{{\left( {\sqrt 2 } \right)}^n} - 1} \right] \to \lim {u_n} = + \infty \) vì \(\left\{{}\begin{matrix}a=2-\sqrt{2}>0\\q=\sqrt{2}>1\end{matrix}\right.\)

Ta có: \(tan\alpha\in\left(0;1\right)\) với mọi \(\alpha \in \left( {0;\dfrac{\pi }{4}} \right) \), do đó:

\(S = \underbrace {1 - \tan \alpha + {{\tan }^2}\alpha - {{\tan }^3}\alpha + ...}_{CSN\_lvh:{u_1} = 1,q = - \tan \alpha } = \dfrac{1}{{1 + \tan \alpha }} = \dfrac{{\cos \alpha }}{{\sin \alpha + \cos \alpha }} = \dfrac{{\cos \alpha }}{{\sqrt 2 \sin \left( {\alpha + \dfrac{\pi }{4}} \right)}}\)

Câu 4.

\(\lim \left( {{n^2}\sin \dfrac{{n\pi }}{5} - 2{n^3}} \right) = \lim {n^3}\left( {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n} - 2} \right) = - \infty \)

Vì \(\lim {n^3} = + \infty ;\lim \left( {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n} - 2} \right) = - 2 \)

\(\left| {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n}} \right| \le \dfrac{1}{n};\lim \dfrac{1}{n} = 0 \Rightarrow \lim \left( {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n} - 2} \right) = - 2\)

Câu 5.

Ta có: \(\left\{ \begin{array}{l} 0 \le \left| {{u_n}} \right| \le \dfrac{1}{{{n^2} + 1}} \le \dfrac{1}{n} \to 0\\ 0 \le \left| {{v_n}} \right| \le \dfrac{1}{{{n^2} + 2}} \le \dfrac{1}{n} \to 0 \end{array} \right. \to \lim {u_n} = \lim {v_n} = 0 \to \lim \left( {{u_n} + {v_n}} \right) = 0\)

Câu 1:

ĐK: $n$ nguyên dương
Ta thấy:

$\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}< \frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2}}+...+\frac{1}{\sqrt{n^2}}=\frac{n}{\sqrt{n^2}}=1$

Và:

$\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}> \frac{1}{\sqrt{n^2+n}}+...+\frac{1}{\sqrt{n^2+n}}=\frac{n}{\sqrt{n^2+n}}$

Trong đó $\lim_{n\to +\infty}\frac{n}{\sqrt{n^2+n}}=\lim_{n\to +\infty}\frac{1}{\sqrt{1+\frac{1}{n}}}=1$

Do đó theo định lý kẹp ta suy ra:

\(\lim_{n\to +\infty}(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}})=1\)

Câu 5:

Ta thấy $\cos 2n$ là hàm bị chặn với mọi $n\to \infty$

$\lim_{n\to \infty}\frac{1}{3n}=0$

$\Rightarrow \lim_{n\to \infty} \sqrt{\frac{\cos 2n}{3n}+9}=\sqrt{0+9}=3$

Câu 3:

\(\lim_{n\to \infty}\sqrt[3]{\frac{5-8n}{n+3}}=\lim_{n\to \infty}\sqrt[3]{\frac{29-8(n+3)}{n+3}}=\lim_{n\to \infty}\sqrt[3]{\frac{29}{n+3}-8}\)

\(=\sqrt[3]{0-8}=-2\)

Câu 4: đã post giải tại bài đăng trước. 

Câu 2:

\(\lim_{n\to \infty}(\sqrt[3]{n^3+1}-n)=\lim_{n\to \infty}\frac{n^3+1-n^3}{\sqrt[3]{(n^3+1)^2}+\sqrt[3]{n^3+1}.n+n^2}=\lim_{n\to \infty}\frac{1}{\sqrt[3]{(n^3+1)^2}+\sqrt[3]{n^3+1}.n+n^2}\)

$=0$

Lời giải:

Có:

\(\frac{\sqrt{2x+1}.\sqrt[3]{3x+1}-1}{x}=\frac{\sqrt{2x+1}(\sqrt[3]{3x+1}-1)+\sqrt{2x+1}-1}{x}=\frac{\sqrt{2x+1}.3x}{x(\sqrt[3]{(3x+1)^2}+\sqrt[3]{3x+1}+1)}+\frac{2x}{x(\sqrt{2x+1}+1)}\)

\(=\frac{3\sqrt{2x+1}}{\sqrt[3]{(3x+1)^2}+\sqrt[3]{3x+1}+1}+\frac{2}{\sqrt{2x+1}+1}\)

Do đó:

\(\lim\limits_{x\to0}\frac{\sqrt{2x+1}.\sqrt[3]{3x+1}-1}{x}=\lim\limits_{x\to0}\frac{3\sqrt{2x+1}}{\sqrt[3]{\left(3x+1\right)^2}+\sqrt[3]{3x+1}+1}+\lim\limits_{x\to0}\frac{2}{\sqrt{2x+1}+1}\)

\(=\frac{3}{1+1+1}+\frac{2}{1+1}=2\)