Bài 2: Giới hạn của hàm số

16.

\(\lim\dfrac{u_n}{v_n}=+\infty\)

17.

\(SA\perp\left(ABC\right)\Rightarrow SA\perp AB\Rightarrow\Delta SAB\) vuông tại A (B đúng)

\(SA\perp\left(ABC\right)\Rightarrow SA\perp BC\) (C đúng)

\(\left\{{}\begin{matrix}SA\perp\left(ABC\right)\Rightarrow SA\perp BC\\AB\perp BC\left(gt\right)\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\) (D đúng)

18.

Tập hợp điểm cách đều 2 điểm AB cho trước là mặt phẳng trung trực của AB

19.

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

11.

Do \(\lim\limits_{x\rightarrow2^-}\left(1-x^2\right)=1-2^2=-3< 0\)

\(\lim\limits_{x\rightarrow2^-}\left(x-2\right)=0\)

Và: \(x-2< 0\) khi \(x< 2\)

\(\Rightarrow\lim\limits_{x\rightarrow2^-}\dfrac{1-x^2}{x-2}=+\infty\)

5.

\(y=\dfrac{2x-1}{1-x}\Rightarrow y'=\dfrac{\left(2x-1\right)'\left(1-x\right)-\left(1-x\right)'\left(2x-1\right)}{\left(1-x\right)^2}\)

\(=\dfrac{2\left(1-x\right)+\left(2x-1\right)}{\left(1-x\right)^2}=\dfrac{1}{\left(1-x\right)^2}=\dfrac{1}{\left(x-1\right)^2}\)

9.

\(\lim\limits\dfrac{2n^2+4}{3-n^2}=\lim\dfrac{2+\dfrac{4}{n^2}}{\dfrac{3}{n^2}-1}=\dfrac{2+0}{0-1}=-2\)

a: \(=\lim\limits_{x\rightarrow2}\dfrac{x^4-4x^2+x^2-4}{x^3-2x^2-x^2+2x+2x-4}\)

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

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x+2\right)\left(x^2+1\right)}{x^2-x+2}=\dfrac{4\left(2^2+1\right)}{2^2-2+2}=5\)

b: \(=\lim\limits_{x\rightarrow-\infty}\left[x\cdot\dfrac{x^2-2-x^2-2}{\sqrt{x^2-2}+\sqrt{x^2+2}}\right]\)

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

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

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

\(=\dfrac{-4}{1+1}=\dfrac{-4}{2}=-2\)

4.

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{25+x}-\sqrt{25-x}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{25+x}-\sqrt{25-x}\right)\left(\sqrt{25+x}+\sqrt{25-x}\right)}{x\left(\sqrt{25+x}+\sqrt{25-x}\right)}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{25+x-\left(25-x\right)}{x\left(\sqrt{25+x}+\sqrt{25-x}\right)}=\lim\limits_{x\rightarrow0}\dfrac{2x}{x\left(\sqrt{25+x}+\sqrt{25-x}\right)}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2}{\sqrt{25+x}+\sqrt{25-x}}=\dfrac{2}{\sqrt{25}+\sqrt{25}}=\dfrac{1}{5}\)

14.

A là khẳng định sai, CD không vuông góc SB

(Vì nếu \(CD\perp SB\)  (1); do \(SA\perp\left(ABCD\right)\Rightarrow SA\perp CD\) (2)

(1);(2) \(\Rightarrow CD\perp\left(SAB\right)\Rightarrow CD\perp AB\) (vô lý do \(CD||AB\))

\(\lim\limits_{x\rightarrow+\infty}\left(ax-\sqrt{x^2+bx+2}\right)=\lim\limits_{x\rightarrow+\infty}x\left(a-\sqrt{1+\dfrac{b}{x}+\dfrac{2}{x^2}}\right)\)

Nếu \(a\ne1\Rightarrow\lim\limits_{x\rightarrow+\infty}\left(a-\sqrt{1+\dfrac{b}{x}+\dfrac{2}{x^2}}\right)=a-1\ne0\)

\(\Rightarrow\lim\limits_{x\rightarrow+\infty}x\left(a-\sqrt{1+\dfrac{b}{x}+\dfrac{2}{x^2}}\right)=\infty\) ko thỏa mãn giả thiết \(=4\) (hữu hạn)

\(\Rightarrow a=1\)

\(\lim\limits_{x\rightarrow+\infty}\left(x-\sqrt{x^2+bx+2}\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{-bx-2}{x+\sqrt{x^2+bx+2}}=\lim\limits_{x\rightarrow+\infty}\dfrac{-b-\dfrac{2}{x}}{1+\sqrt{1+\dfrac{b}{x}+\dfrac{2}{x^2}}}=-\dfrac{b}{2}\)

\(\Rightarrow-\dfrac{b}{2}=4\Rightarrow b=-8\)

7.

\(y'=\left(-x^3\right)'-\left(5x\right)'+\left(2\right)'=-3x^2-5\)

\(y''=\left(-3x^2\right)'-\left(5\right)'=-6x\)

8.

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

Do: \(\lim\limits_{x\rightarrow+\infty}x^3=+\infty\)

\(\lim\limits_{x\rightarrow+\infty}\left(1+\dfrac{3}{x^2}-\dfrac{2}{x^3}\right)=1>0\)

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

19. Giới hạn đã cho hữu hạn khi và chỉ khi \(a=1\)

Khi đó:

\(\lim\limits_{x\rightarrow+\infty}\left(x-\sqrt{x^2+bx-2}\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-\left(x^2+bx-2\right)}{x+\sqrt{x^2+bx-2}}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{-bx+2}{x+\sqrt{x^2+bx-2}}=\lim\limits_{x\rightarrow+\infty}\dfrac{-b+\dfrac{2}{x}}{1+\sqrt{1+\dfrac{b}{x}-\dfrac{2}{x^2}}}=-\dfrac{b}{2}\)

\(\Rightarrow-\dfrac{b}{2}=3\Rightarrow b=-6\Rightarrow a+b=1+\left(-6\right)=-5\)

20.

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