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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\lim\limits_{x\rightarrow1}f\left(x\right)=\lim\limits_{x\rightarrow1}\dfrac{2x^2-5x+3}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(2x-3\right)}{x-1}=\lim\limits_{x\rightarrow1}2x-3=2\cdot1-3=-1\)
f(1)=4
=>\(\lim\limits_{x\rightarrow1}f\left(x\right)< >f\left(1\right)\)
=>Hàm số bị gián đoạn tại x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}x^2+3x+1=1+3\cdot1+1=5\)
\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}2x+2=2\cdot1+2=4\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
Đề lỗi công thức toán rồi bạn. Không nhìn thấy được biểu thức hiển thị.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\lim\limits_{x\rightarrow2}f\left(x\right)=\lim\limits_{x\rightarrow2}\dfrac{2-\sqrt{2x^2-4}}{2-x}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{4-2x^2+4}{2+\sqrt{2x^2-4}}\cdot\dfrac{1}{2-x}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{-2\left(x^2-4\right)}{-\left(x-2\right)\left(2+\sqrt{2x^2-4}\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(2+\sqrt{2x^2-4}\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{2\left(x+2\right)}{2+\sqrt{2x^2-4}}=\dfrac{2\left(2+2\right)}{2+\sqrt{2\cdot2^2-4}}\)
\(=\dfrac{2\cdot4}{2+2}=\dfrac{8}{4}=2\)
\(f\left(2\right)=1\)
=>\(\lim\limits_{x\rightarrow2}f\left(x\right)< >f\left(2\right)\)
=>Hàm số bị gián đoạn tại x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\lim\limits_{x\rightarrow5}f\left(x\right)=\lim\limits_{x\rightarrow5}\dfrac{x^2-8x+15}{x-5}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\left(x-3\right)\left(x-5\right)}{x-5}=\lim\limits_{x\rightarrow5}x-3=5-3=2\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x+3}-2}{x-1}=\lim\limits_{x\rightarrow1^+}\dfrac{x-1}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}=\lim\limits_{x\rightarrow1^+}\dfrac{1}{\sqrt{x+3}+2}=\dfrac{1}{4}\)
\(f\left(1\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\left(ax+2\right)=a+2\)
Hàm liên tục tại x=1 khi:
\(a+2=\dfrac{1}{4}\Rightarrow a=-\dfrac{7}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x+3}-2}{x-1}=\lim\limits_{x\rightarrow1^+}\dfrac{x-1}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}=\lim\limits_{x\rightarrow1^+}\dfrac{1}{\sqrt{x+3}+2}=\dfrac{1}{4}\)
\(f\left(1\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\left(mx\right)=m\)
Hàm liên tục tại x=1 khi: \(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=f\left(1\right)\)
\(\Leftrightarrow m=\dfrac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\lim\limits_{x\rightarrow2}f\left(x\right)=\lim\limits_{x\rightarrow2}\dfrac{2x^2-7x+6}{2-x}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{2x^2-4x-3x+6}{-\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(3-2x\right)}{x-2}=\lim\limits_{x\rightarrow2}3-2x=3-2\cdot2=3-4=-1\)
\(f\left(2\right)=2\cdot2-5=-1\)
=>\(\lim\limits_{x\rightarrow2}f\left(x\right)=f\left(2\right)\)
=>Hàm số liên tục tại x=2
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