1.a
\(\lim\dfrac{3n^3+2n^2+n}{n^3+4}=\lim\dfrac{n^3\left(3+\dfrac{2}{n}+\dfrac{1}{n^2}\right)}{n^3\left(1+\dfrac{4}{n^3}\right)}\)
\(=\lim\dfrac{3+\dfrac{2}{n}+\dfrac{1}{n^2}}{1+\dfrac{4}{n^3}}=\dfrac{3+0+0}{1+0}=3\)
b.
\(\lim\limits_{x\rightarrow3}\dfrac{x^2+2x-15}{x-3}=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)\left(x+5\right)}{x-3}\)
\(=\lim\limits_{x\rightarrow3}\left(x+5\right)=8\)
2.
Ta có:
\(\lim\limits_{x\rightarrow5}f\left(x\right)=\lim\limits_{x\rightarrow5}\dfrac{x^2-25}{x-5}=\lim\limits_{x\rightarrow5}\dfrac{\left(x-5\right)\left(x+5\right)}{x-5}\)
\(=\lim\limits_{x\rightarrow5}\left(x+5\right)=10\)
Và: \(f\left(5\right)=9\)
\(\Rightarrow\lim\limits_{x\rightarrow5}f\left(x\right)\ne f\left(5\right)\)
\(\Rightarrow\) Hàm gián đoạn tại \(x_0=5\)
3.
\(\lim\limits_{x\rightarrow1}\dfrac{3x^2-2x-1}{x^3-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(3x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{3x+1}{x^2+x+1}=\dfrac{3.1+1}{1^2+1+1}=\dfrac{4}{3}\)
b.
\(\lim\limits_{x\rightarrow3^-}\dfrac{x+3}{x-3}\)
Do: \(\lim\limits_{x\rightarrow3^-}\left(x+3\right)=6>0\)
\(\lim\limits_{x\rightarrow3^-}\left(x-3\right)=0\)
Và \(x-3< 0\) khi \(x< 3\)
\(\Rightarrow\lim\limits_{x\rightarrow3^-}\dfrac{x+3}{x-3}=-\infty\)
4.
Ta có:
\(\lim\limits_{x\rightarrow2}f\left(x\right)=\lim\limits_{x\rightarrow2}\dfrac{x-2}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{1}{x+2}=\dfrac{1}{4}\)
Và: \(f\left(2\right)=\dfrac{3}{2}\)
\(\Rightarrow\lim\limits_{x\rightarrow2}f\left(x\right)\ne f\left(2\right)\)
\(\Rightarrow\) Hàm gián đoạn tại \(x_0=2\)
5.
a. Do ABCD.A'B'C'D' là hình hộp \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{B'C'}=\overrightarrow{BC}\\\overrightarrow{DD'}=\overrightarrow{CC'}\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{AB}+\overrightarrow{B'C'}+\overrightarrow{DD'}=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CC'}=\overrightarrow{AC}+\overrightarrow{CC'}=\overrightarrow{AC'}\) (đpcm)
b.
\(\overrightarrow{BD}-\overrightarrow{D'D}-\overrightarrow{B'D'}=\overrightarrow{BD}+\overrightarrow{DD'}+\overrightarrow{D'B'}=\overrightarrow{BD'}+\overrightarrow{D'B'}=\overrightarrow{BB'}\)
c.
Do ABCD.A'B'C'D' là hình hộp \(\Rightarrow\overrightarrow{C'A'}=\overrightarrow{CA}\)
Do đó:
\(\overrightarrow{AC}+\overrightarrow{BA'}+\overrightarrow{DB}+\overrightarrow{C'D}=\overrightarrow{AC}+\left(\overrightarrow{DB}+\overrightarrow{BA'}\right)+\overrightarrow{C'D}\)
\(=\overrightarrow{AC}+\overrightarrow{DA'}+\overrightarrow{C'D}=\overrightarrow{AC}+\left(\overrightarrow{C'D}+\overrightarrow{DA'}\right)=\overrightarrow{AC}+\overrightarrow{C'A'}\)
\(=\overrightarrow{AC}+\overrightarrow{CA}=\overrightarrow{0}\)
Ai giúp em câu này với ạ :((