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a/ \(\lim\limits_{x\rightarrow\sqrt{2}}f\left(x\right)=\lim\limits_{x\rightarrow\sqrt{2}}\frac{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}{x-\sqrt{2}}=\lim\limits_{x\rightarrow\sqrt{2}}\left(x+\sqrt{2}\right)=2\sqrt{2}\)
\(\Rightarrow\lim\limits_{x\rightarrow\sqrt{2}}f\left(x\right)=f\left(\sqrt{2}\right)\Rightarrow\) hàm số liên tục tại \(x=\sqrt{2}\)
b/ \(\lim\limits_{x\rightarrow5^+}f\left(x\right)=\lim\limits_{x\rightarrow5^+}\frac{x-5}{\sqrt{2x-1}-3}=\frac{\left(x-5\right)\left(\sqrt{2x-1}+3\right)}{2\left(x-5\right)}=\lim\limits_{x\rightarrow5^+}\frac{\sqrt{2x-1}+3}{2}=3\)
\(f\left(5\right)=\lim\limits_{x\rightarrow5^-}f\left(x\right)=\lim\limits_{x\rightarrow5^-}\left[\left(x-5\right)^2+3\right]=5\)
\(\Rightarrow\lim\limits_{x\rightarrow5^+}f\left(x\right)=\lim\limits_{x\rightarrow5^-}f\left(x\right)=f\left(5\right)\Rightarrow\) hàm số liên tục tại \(x=5\)
\(\lim\limits_{x\rightarrow5}f\left(x\right)=\lim\limits_{x\rightarrow5}\dfrac{\sqrt{2x-9}-1}{5-x}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{2x-9-1}{\sqrt{2x-9}+1}\cdot\dfrac{1}{5-x}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{2\left(x-5\right)}{-\left(x-5\right)\left(\sqrt{2x-9}+1\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{-2}{\sqrt{2x-9+1}}=\dfrac{-2}{\sqrt{10-9}+1}=-\dfrac{2}{2}=-1\)
f(5)=3
=>\(\lim\limits_{x\rightarrow5}f\left(x\right)< >f\left(5\right)\)
=>Hàm số bị gián đoạn tại x=5
\(\lim\limits_{x\rightarrow2}f\left(x\right)=\lim\limits_{x\rightarrow2}\dfrac{2-\sqrt{2x^2-4}}{2-x}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{4-2x^2+4}{2+\sqrt{2x^2-4}}\cdot\dfrac{1}{2-x}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{-2\left(x^2-4\right)}{-\left(x-2\right)\left(2+\sqrt{2x^2-4}\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(2+\sqrt{2x^2-4}\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{2\left(x+2\right)}{2+\sqrt{2x^2-4}}=\dfrac{2\left(2+2\right)}{2+\sqrt{2\cdot2^2-4}}\)
\(=\dfrac{2\cdot4}{2+2}=\dfrac{8}{4}=2\)
\(f\left(2\right)=1\)
=>\(\lim\limits_{x\rightarrow2}f\left(x\right)< >f\left(2\right)\)
=>Hàm số bị gián đoạn tại x=2
\(\lim\limits_{x\rightarrow5}f\left(x\right)=\lim\limits_{x\rightarrow5}\dfrac{x^2-8x+15}{x-5}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\left(x-3\right)\left(x-5\right)}{x-5}=\lim\limits_{x\rightarrow5}x-3=5-3=2\)
f(5)=2*5-1=9
=>\(f\left(5\right)\ne\lim\limits_{x\rightarrow5}f\left(x\right)\)
=>Hàm số gián đoạn tại x=5
\(\lim\limits_{x\rightarrow2}f\left(x\right)=\lim\limits_{x\rightarrow2}\dfrac{2x^2-7x+6}{2-x}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{2x^2-4x-3x+6}{-\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(3-2x\right)}{x-2}=\lim\limits_{x\rightarrow2}3-2x=3-2\cdot2=3-4=-1\)
\(f\left(2\right)=2\cdot2-5=-1\)
=>\(\lim\limits_{x\rightarrow2}f\left(x\right)=f\left(2\right)\)
=>Hàm số liên tục tại x=2
\(\lim\limits_{x\rightarrow-2}f\left(x\right)=\dfrac{2x^2-x-10}{x+2}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{2x^2+4x-5x-10}{x+2}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{\left(x+2\right)\left(2x-5\right)}{x+2}\)
\(=\lim\limits_{x\rightarrow-2}2x-5=2\cdot\left(-2\right)-5=-9\)
\(f\left(-2\right)=a-2\)
hàm số liên tục tại x=-2 khi a-2=-9
=>a=-7
Hàm số không liên tục tại x=-2 thì \(a-2\ne-9\)
=>\(a\ne-7\)
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}x^2+3x+1=1+3\cdot1+1=5\)
\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}2x+2=2\cdot1+2=4\)
f(1)=1+3+1=5
=>\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=f\left(1\right)\ne\lim\limits_{x\rightarrow1^-}f\left(x\right)\)
=>Hàm số bị gián đoạn tại x=1
