\(\lim\limits_{x\rightarrow2}f\left(x\right)=\lim\limits_{x\rightarrow2}\dfrac{2-\sqrt{2x^2-4}}{2-x}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{4-2x^2+4}{2+\sqrt{2x^2-4}}\cdot\dfrac{1}{2-x}\)

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

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

\(=\lim\limits_{x\rightarrow2}\dfrac{2\left(x+2\right)}{2+\sqrt{2x^2-4}}=\dfrac{2\left(2+2\right)}{2+\sqrt{2\cdot2^2-4}}\)

\(=\dfrac{2\cdot4}{2+2}=\dfrac{8}{4}=2\)

\(f\left(2\right)=1\)

=>\(\lim\limits_{x\rightarrow2}f\left(x\right)< >f\left(2\right)\)

=>Hàm số bị gián đoạn tại x=2

\(\lim\limits_{x\rightarrow2}f\left(x\right)=\lim\limits_{x\rightarrow2}\dfrac{2x^2-7x+6}{2-x}\)

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

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

\(f\left(2\right)=2\cdot2-5=-1\)

=>\(\lim\limits_{x\rightarrow2}f\left(x\right)=f\left(2\right)\)

=>Hàm số liên tục tại x=2

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

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

\(=-\left(2+2\right)=-4\)

\(f\left(2\right)=2-7=-5\)

=>\(\lim\limits_{x\rightarrow2}f\left(x\right)< >f\left(2\right)\)

=>Hàm số gián đoạn tại x=2

Khi \(x\ne\)2 thì \(f\left(x\right)=\dfrac{x^2-4}{2-x}\) hoàn toàn xác định nên hàm số liên tục trên các khoảng \(\left(-\infty;2\right);\left(2;+\infty\right)\)

\(\lim\limits_{x\rightarrow5}f\left(x\right)=\lim\limits_{x\rightarrow5}\dfrac{\sqrt{2x-9}-1}{5-x}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{2x-9-1}{\sqrt{2x-9}+1}\cdot\dfrac{1}{5-x}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{2\left(x-5\right)}{-\left(x-5\right)\left(\sqrt{2x-9}+1\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{-2}{\sqrt{2x-9+1}}=\dfrac{-2}{\sqrt{10-9}+1}=-\dfrac{2}{2}=-1\)

f(5)=3

=>\(\lim\limits_{x\rightarrow5}f\left(x\right)< >f\left(5\right)\)

=>Hàm số bị gián đoạn tại x=5

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

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(2x-3\right)}{x-1}=\lim\limits_{x\rightarrow1}2x-3=2\cdot1-3=-1\)

f(1)=4

=>\(\lim\limits_{x\rightarrow1}f\left(x\right)< >f\left(1\right)\)

=>Hàm số bị gián đoạn tại x=1

Lời giải:
\(\lim\limits_{x\to 1+}f(x)=\lim\limits_{x\to 1+}(5x-2)=3\)

\(\lim \limits_{x\to 1-}f(x)=\lim \limits_{x\to 1-}(2+2x)=4\)

\(\Rightarrow \lim\limits_{x\to 1+}f(x)\neq \lim \limits_{x\to 1-}f(x)\)

Do đó hàm số không liên tục tại $x=1$

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}2x+2=2\cdot1+2=4\)

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

\(f\left(1\right)=2+2\cdot2=4\)

Vì \(\lim\limits_{x\rightarrow1^-}f\left(x\right)< >\lim\limits_{x\rightarrow1^+}f\left(x\right)\)

nên hàm số bị gián đoạn tại x=1

\(\lim\limits_{x\rightarrow-2}f\left(x\right)=\dfrac{2x^2-x-10}{x+2}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{2x^2+4x-5x-10}{x+2}\)

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

\(=\lim\limits_{x\rightarrow-2}2x-5=2\cdot\left(-2\right)-5=-9\)

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

hàm số liên tục tại x=-2 khi a-2=-9

=>a=-7

Hàm số không liên tục tại x=-2 thì \(a-2\ne-9\)

=>\(a\ne-7\)