Question #305565

Determine whether if lim f(c) = f(c)


x + c



1. f(x) = x+2; c = -1



2. f(x) = x-2; c = 0



3. (at c = -1 )


f(x) = {x ² - 1 if x = < -1}


f(x) = { (x - 1) ² - 4 if x = ≥ - 1}



4. (at c = 1 )


f(x) = {x³ - 1 if x = < 1}


f(x) = { x² +4 if x = ≥ 1}

1
Expert's answer
2022-03-04T12:25:17-0500

1, The function f(x)=x+2f(x)=x+2 is continuous on R\R as a polynomial.

Therefore


limxcf(x)=f(c),cR\lim\limits_{x\to c}f(x)=f(c), c\in \R

Then


limx1f(x)=f(1)=1+2=1\lim\limits_{x\to -1}f(x)=f(-1)=-1+2=1


2, The function f(x)=x2f(x)=x-2 is continuous on R\R as a polynomial.

Therefore


limxcf(x)=f(c),cR\lim\limits_{x\to c}f(x)=f(c), c\in \R

Then


limx0f(x)=f(0)=02=2\lim\limits_{x\to0}f(x)=f(0)=0-2=-2



3.


limx1f(x)=(1)21=0\lim\limits_{x\to -1^-}f(x)=(-1)^2-1=0

limx1+f(x)=(11)24=0\lim\limits_{x\to -1^+}f(x)=(-1-1)^2-4=0

Since


limx1f(x)=0=limx1+f(x),\lim\limits_{x\to -1^-}f(x)=0=\lim\limits_{x\to -1^+}f(x),

then limx1f(x)\lim\limits_{x\to -1}f(x) exists, and limx1f(x)=0.\lim\limits_{x\to -1}f(x)=0.


4.


limx1f(x)=(1)31=0\lim\limits_{x\to 1^-}f(x)=(1)^3-1=0

limx1+f(x)=(1)2+4=5\lim\limits_{x\to 1^+}f(x)=(1)^2+4=5

Since


limx1f(x)=05=limx1+f(x),\lim\limits_{x\to 1^-}f(x)=0\not=5=\lim\limits_{x\to 1^+}f(x),

then limx1f(x)\lim\limits_{x\to 1}f(x) does not exist.



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