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

\(=\dfrac{-2-5}{\left(-2\right)^2+4}=\dfrac{-7}{4+4}=-\dfrac{7}{8}\)

\(\lim\limits_{x\rightarrow2^+}f\left(x\right)=\lim\limits_{x\rightarrow2^+}\dfrac{x^2-3x}{x-1}=\dfrac{2^2-3\cdot2}{2-1}=\dfrac{4-6}{1}=-2\)

\(f\left(2\right)=3a-5\)

\(\lim\limits_{x\rightarrow2^-}f\left(x\right)=\lim\limits_{x\rightarrow2^-}3a-5\)

Để \(\lim\limits_{x\rightarrow2}f\left(x\right)\) tồn tại thì 3a-5=-2

=>3a=3

=>a=1

ĐKXĐ: \(b>=0\)

Để \(-\dfrac{1}{\sqrt{2}};\sqrt{b};\sqrt{2}\) theo thứ tự lập được thành cấp số nhân thì \(\left(\sqrt{b}\right)^2=-\dfrac{1}{\sqrt{2}}\cdot\sqrt{2}\)

=>b=-1(loại)

=>\(b\in\varnothing\)

\(u_8=u_5.q^{8-5}\Rightarrow-16=2.q^3\Rightarrow q=-2\)

\(u_5=u_1.q^4\Rightarrow u_1=\dfrac{u_5}{q^4}=\dfrac{2}{\left(-2\right)^4}=\dfrac{1}{8}\)

\(u_{21}=u_1.q^{20}=\dfrac{1}{8}.\left(-2\right)^{20}=2^{17}\)

\(\left\{{}\begin{matrix}u_2+u_5-u_4=10\left(1\right)\\u_3+u_6-u_5=20\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u_2+u_5-u_4=10\\u_2.q+u_5.q-u_4.q=20\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u_2+u_5-u_4=10\\q.\left(u_2+u_5-u_4\right)=20\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u_2+u_5-u_4=10\\q.10=20\end{matrix}\right.\)

\(\Rightarrow q=2\)

\(\left(1\right)\Rightarrow u_1.q+u_1.q^4-u_1.q^3=10\)

\(\Leftrightarrow u_1.\left(q+q^4-q^3\right)=10\)

\(\Leftrightarrow u_1.=\dfrac{10}{\left(q+q^4-q^3\right)}=\dfrac{10}{2+2^4-2^3}=1\)

Vậy \(u_1=1;q=2\)

\(f\left(x\right)=\dfrac{\sqrt{x^2-3x+9}-\left(2x-3\right)}{2x-6}\)

\(=\dfrac{x^2-3x+9-4x^2+12x-9}{2\left(x-3\right)\left(\sqrt{x^2-3x+9}+\left(2x-3\right)\right)}\)

\(=\dfrac{-3x^2+9x}{2\left(x-3\right)\left(\sqrt{x^2-3x+9}+\left(2x-3\right)\right)}\)

\(=\dfrac{-3x\left(x-3\right)}{2\left(x-3\right)\left(\sqrt{x^2-3x+9}+\left(2x-3\right)\right)}\)

\(=\dfrac{-3x}{2\left(\sqrt{x^2-3x+9}+\left(2x-3\right)\right)}\)

\(\lim\limits_{x\rightarrow3}f\left(x\right)=\dfrac{-3.3}{2\left(\sqrt{9-9+9}+\left(6-3\right)\right)}=-\dfrac{3}{4}\)

\(...\Rightarrow u_n=\dfrac{1}{2}.\dfrac{2n^2-3+\dfrac{5}{2}}{2n^2-3}=\dfrac{1}{2}+\dfrac{5}{2\left(2n^2-3\right)}\)

Ta thấy \(\dfrac{1}{2}+\dfrac{5}{2.\left(-3\right)}\le u_n\le\dfrac{1}{2}+0\left(n\ge0\right)\)

\(\Rightarrow-\dfrac{1}{3}\le u_n\le\dfrac{1}{2}\)

\(\Rightarrow u_n\) bị chặn trên bởi \(\dfrac{1}{2}\) và bị chặn dưới bởi \(-\dfrac{1}{3}\)

\(f\left(1\right)=\dfrac{3.1^2-5}{1+2}=-\dfrac{2}{3}\)

\(\lim\limits_{x\rightarrow1}f\left(x\right)=\lim\limits_{x\rightarrow1}\dfrac{3x^2-5}{x+2}=\dfrac{3.1^2-5}{1+2}=-\dfrac{2}{3}\)

\(\Rightarrow f\left(1\right)=\lim\limits_{x\rightarrow1}f\left(x\right)=-\dfrac{2}{3}\)

\(\Rightarrow\) Hàm số \(f\left(x\right)\) liên tục tại \(x=1\)

\(\lim\limits_{x\rightarrow6+}f\left(x\right)=\lim\limits_{x\rightarrow6+}\dfrac{x^2-36}{x-6}=\lim\limits_{x\rightarrow6+}x+6=12\)

\(\lim\limits_{x\rightarrow6-}f\left(x\right)=\lim\limits_{x\rightarrow6-}4x-12=12\)

\(\Rightarrow\lim\limits_{x\rightarrow6}f\left(x\right)=\lim\limits_{x\rightarrow6+}f\left(x\right)=\lim\limits_{x\rightarrow6-}f\left(x\right)=12\)