f ( x ) = { 1 + 2 x , x ≤ 0 3 x − 2 , 0 < x ≤ 1 2 x 2 − 1 , x > 1 f(x)=\begin{cases}
1+2x,\qquad x\leq 0
\\
3x-2,\qquad 0<x\leq 1
\\2x^2-1,\qquad x>1
\end{cases} f ( x ) = ⎩ ⎨ ⎧ 1 + 2 x , x ≤ 0 3 x − 2 , 0 < x ≤ 1 2 x 2 − 1 , x > 1
1) x = 0 x=0 x = 0
lim x → 0 − 0 f ( x ) = lim x → 0 − 0 ( 1 + 2 x ) = 1 \lim\limits_{x\rightarrow 0-0}f( x)= \lim\limits_{x\rightarrow 0-0}(1+2x)=1 x → 0 − 0 lim f ( x ) = x → 0 − 0 lim ( 1 + 2 x ) = 1
lim x → 0 + 0 f ( x ) = lim x → 0 + 0 ( 3 x − 2 ) = − 2 \lim\limits_{x\rightarrow 0+0}f( x)= \lim\limits_{x\rightarrow 0+0}(3x-2)=-2 x → 0 + 0 lim f ( x ) = x → 0 + 0 lim ( 3 x − 2 ) = − 2
lim x → 0 − 0 f ( x ) ≠ lim x → 0 + 0 f ( x ) \lim\limits_{x\rightarrow 0-0}f(x)\neq \lim\limits_{x\rightarrow 0+0}f(x) x → 0 − 0 lim f ( x ) = x → 0 + 0 lim f ( x )
f ( x ) f(x) f ( x ) is discontinuous at x = 0 x=0 x = 0
2) x = 1 x=1 x = 1
lim x → 1 − 0 f ( x ) = lim x → 1 − 0 ( 3 x − 2 ) = 1 \lim\limits_{x\rightarrow 1-0}f( x)= \lim\limits_{x\rightarrow 1-0}(3x-2)=1 x → 1 − 0 lim f ( x ) = x → 1 − 0 lim ( 3 x − 2 ) = 1
lim x → 1 + 0 f ( x ) = lim x → 1 + 0 ( 2 x 2 − 1 ) = 1 \lim\limits_{x\rightarrow 1+0}f( x)= \lim\limits_{x\rightarrow 1+0}(2x^2-1)=1 x → 1 + 0 lim f ( x ) = x → 1 + 0 lim ( 2 x 2 − 1 ) = 1
lim x → 1 − 0 f ( x ) = lim x → 1 + 0 f ( x ) = f ( 1 ) \lim\limits_{x\rightarrow 1-0}f(x)=\lim\limits_{x\rightarrow 1+0}f(x)=f(1) x → 1 − 0 lim f ( x ) = x → 1 + 0 lim f ( x ) = f ( 1 )
f ( x ) f(x) f ( x ) is continuous at x = 1 x=1 x = 1
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