1.
\(\lim\left(\sqrt{9^n-2.3^n}-3^n+\dfrac{1}{2021}\right)\)
\(=\lim\left(\dfrac{\left(\sqrt{9^n-2.3^n}-3^n\right)\left(\sqrt{9^n-2.3^n}+3^n\right)}{\sqrt{9^n-2.3^n}+3^n}+\dfrac{1}{2021}\right)\)
\(=\lim\left(\dfrac{-2.3^n}{\sqrt{9^n-2.3^n}+3^n}+\dfrac{1}{2021}\right)\)
\(=\lim\left(\dfrac{-2.3^n}{3^n\left(\sqrt{1-\dfrac{2}{3^n}}+1\right)}+\dfrac{1}{2021}\right)\)
\(=\lim\left(\dfrac{-2}{\sqrt{1-\dfrac{2}{3^n}}+1}+\dfrac{1}{2021}\right)\)
\(=\dfrac{-2}{1+1}+\dfrac{1}{2021}=-\dfrac{2020}{2021}\)
2.
\(AP=4PB=4\left(AB-AP\right)=4AB-4AP\)
\(\Rightarrow5AP=4AB\Rightarrow AP=\dfrac{4}{5}AB\)
\(\Rightarrow\overrightarrow{AP}=\dfrac{4}{5}\overrightarrow{AB}\)
\(CD=5CQ=5\left(CD-DQ\right)\Rightarrow5DQ=4CD\Rightarrow DQ=\dfrac{4}{5}CD\)
\(\Rightarrow\overrightarrow{DQ}=-\dfrac{4}{5}\overrightarrow{CD}\)
Ta có:
\(\overrightarrow{PQ}=\overrightarrow{PA}+\overrightarrow{AD}+\overrightarrow{DQ}=-\dfrac{4}{5}\overrightarrow{AB}+\overrightarrow{AD}-\dfrac{4}{5}\overrightarrow{CD}\)
\(=-\dfrac{4}{5}\left(\overrightarrow{AD}+\overrightarrow{DB}\right)+\overrightarrow{AD}-\dfrac{4}{5}\overrightarrow{CD}=-\dfrac{4}{5}\overrightarrow{AD}-\dfrac{4}{5}\overrightarrow{DB}+\overrightarrow{AD}-\dfrac{4}{5}\overrightarrow{CD}\)
\(=\dfrac{1}{5}\overrightarrow{AD}-\dfrac{4}{5}\left(\overrightarrow{CD}+\overrightarrow{DB}\right)=\dfrac{1}{5}\overrightarrow{AD}-\dfrac{4}{5}\overrightarrow{CB}\)
\(=\dfrac{1}{5}\overrightarrow{AD}+\dfrac{4}{5}\overrightarrow{BC}\)
Mà \(\overrightarrow{AD};\overrightarrow{BC}\) không cùng phương\(\Rightarrow\overrightarrow{AD};\overrightarrow{BC};\overrightarrow{PQ}\) đồng phẳng
3.
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[]{2x-1}-\sqrt[3]{3x-2}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[]{2x-1}-1+1-\sqrt[3]{3x-2}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{\left(\sqrt[]{2x-1}-1\right)\left(\sqrt[]{2x-1}+1\right)}{\sqrt[]{2x-1}+1}+\dfrac{\left(1-\sqrt[3]{3x-2}\right)\left(1+\sqrt[3]{3x-2}+\sqrt[3]{\left(3x-2\right)^2}\right)}{1+\sqrt[3]{3x-2}+\sqrt[3]{\left(3x-2\right)^2}}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{2x-1-1}{\sqrt[]{2x-1}+1}+\dfrac{1-\left(3x-2\right)}{1+\sqrt[3]{3x-2}+\sqrt[3]{\left(3x-2\right)^2}}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{2\left(x-1\right)}{\sqrt[]{2x-1}+1}-\dfrac{3\left(x-1\right)}{1+\sqrt[3]{3x-2}+\sqrt[3]{\left(3x-2\right)^2}}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\sqrt[]{2x-1}+1}-\dfrac{3}{1+\sqrt[3]{3x-2}+\sqrt[3]{\left(3x-2\right)^2}}\right)\)
\(=\dfrac{2}{1+1}-\dfrac{3}{1+1+1}=0\)
4.
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt[]{3x+1}-2}{x^2-1}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{\left(\sqrt[]{3x+1}-2\right)\left(\sqrt[]{3x+1}+2\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt[]{3x+1}+2\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt[]{3x+1}+2\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{3}{\left(x+1\right)\left(\sqrt[]{3x+1}+2\right)}\)
\(=\dfrac{3}{2\left(2+2\right)}=\dfrac{3}{8}\)
\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\dfrac{a\left(x^2-2\right)}{x-3}=\dfrac{a\left(1-2\right)}{1-3}=\dfrac{a}{2}\)
\(f\left(1\right)=\dfrac{a\left(1-2\right)}{1-3}=\dfrac{a}{2}\)
Hàm liên tục tại \(x=1\) khi và chỉ khi:
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=f\left(1\right)\Leftrightarrow\dfrac{a}{2}=\dfrac{3}{8}\)
\(\Rightarrow a=\dfrac{3}{4}\)
