Bài 1: Giới hạn của dãy số

7.

Hàm có đúng 1 điểm gián đoạn khi và chỉ khi \(x^2-2\left(m+2\right)x+4=0\) có đúng 1 nghiệm

\(\Rightarrow\Delta'=\left(m+2\right)^2-4=0\)

\(\Leftrightarrow m^2+4m=0\Rightarrow\left[{}\begin{matrix}m=-4\\m=0\end{matrix}\right.\)

\(-4+0=-4\)

8.

Hàm gián đoạn khi \(x^2+2x-3=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

Nên hàm đồng biến trên các khoảng \(\left(-\infty;-3\right);\left(-3;1\right);\left(1;+\infty\right)\) và các tập con của chúng

A đúng

20: \(\lim\limits_{x\rightarrow+\infty}x^3+2x-1=\lim\limits_{x\rightarrow+\infty}\left[x^3\left(1+\dfrac{2}{x^2}-\dfrac{1}{x^3}\right)\right]\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow+\infty}x^3=+\infty\\\lim\limits_{x\rightarrow+\infty}1+\dfrac{2}{x^2}-\dfrac{1}{x^3}=1\end{matrix}\right.\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{3x^2+2x-1}-2}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{3x^2+2x-5}{\left(x-1\right)\left(x+1\right)\left(\sqrt{3x^2+2x-1}+2\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(3x+5\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{3x^2+2x-1}+2\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{3x+5}{\left(x+1\right)\left(\sqrt{3x^2+2x-1}+2\right)}=\dfrac{8}{2\left(\sqrt{4}+2\right)}=1\)

\(\lim\limits_{x\rightarrow2}\dfrac{x^3+x^2-12}{x^2-3x+2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x^2+3x+6\right)}{\left(x-2\right)\left(x-1\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+3x+6}{x-1}=\dfrac{4+6+6}{1}=16\)

\(\lim\limits_{x\rightarrow1}\dfrac{x^3+3x^2-4}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x^2+4x+4\right)}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\left(x+2\right)^2\)

\(=9\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{4x+5}+x}{x^2+3x+2}=\dfrac{\sqrt{4.1+5}+1}{1^2+3.1+2}=\dfrac{2}{3}\)

(Đề là \(x\rightarrow-1\) thì hợp lý hơn)

\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{4x+5}+x}{x^2+3x+2}=\lim\limits_{x\rightarrow-1}\dfrac{\left(\sqrt{4x+5}+x\right)\left(\sqrt{4x+5}-x\right)}{\left(x+1\right)\left(x+2\right)\left(\sqrt{4x+5}-x\right)}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{-x^2+4x+5}{\left(x+1\right)\left(x+2\right)\left(\sqrt{4x+5}-x\right)}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(5-x\right)}{\left(x+1\right)\left(x+2\right)\left(\sqrt{4x+5}-x\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{5-x}{\left(x+2\right)\left(\sqrt{4x+5}-x\right)}=\dfrac{6}{1\left(1+1\right)}=3\)

\(\lim\limits_{x\rightarrow-3}\dfrac{x^3+27}{2x^2+3x-9}=\lim\limits_{x\rightarrow-3}\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{\left(x+3\right)\left(2x-3\right)}=\lim\limits_{x\rightarrow-3}\dfrac{x^2-3x+9}{2x-3}=-3\)

\(lim_{x\rightarrow1}\dfrac{3-2x^2-x^4}{5-3x-2x^2}=\)\(lim_{x\rightarrow1}\dfrac{\left(x^2-1\right)\left(x^2+3\right)}{2\left(x-1\right)\left(x+\dfrac{5}{2}\right)}=\)\(lim_{x\rightarrow1}\dfrac{\left(x+1\right)\left(x^2+3\right)}{2\left(x+\dfrac{5}{2}\right)}=\dfrac{\left(1+1\right)\left(1+3\right)}{2\left(1+\dfrac{5}{2}\right)}=\dfrac{8}{7}\)

\(\lim\left(2n+1-\sqrt{4n^2-3}\right)=\lim\dfrac{\left(2n+1\right)^2-\left(4n^2-3\right)}{2n+1+\sqrt{4n^2-3}}\)

\(=\lim\dfrac{4n+4}{2n+1+\sqrt{4n^2-3}}=\lim\dfrac{4+\dfrac{4}{n}}{2+\dfrac{1}{n}+\sqrt{4-\dfrac{3}{n^2}}}=\dfrac{4}{2+\sqrt{4}}=1\)