Question

if ( f(x) = llbracket x
rbracket + llbracket -x
rbracket ), show that ( limlimits_{x \to 2} f(x) ) exists but is not equal to ( f(2) ).

the graph of ( f(x) = llbracket x
rbracket + llbracket -x
rbracket ) is the same as the graph of ( g(x) = -1 ) with holes at each integer, since ( f(a) = \boxed{0} ) for any integer ( a ). also,

( limlimits_{x \to 2^-} f(x) = \boxed{-1} ) and ( limlimits_{x \to 2^+} f(x) = \boxed{-1} ), so ( limlimits_{x \to 2} f(x) = \boxed{-1} ).

however,

( f(2) = llbracket 2
rbracket + llbracket -2
rbracket = 2 + \boxed{} = \boxed{} ),

so ( limlimits_{x \to 2} f(x)
eq f(2) ).

Explanation:

Step1: Analyze \( f(a) \) for integer \( a \)

For any integer \( a \), \( \llbracket a
rbracket = a \) and \( \llbracket -a
rbracket = -a \). So \( f(a)=\llbracket a
rbracket+\llbracket -a
rbracket=a + (-a)=0 \).

Step2: Find left - hand limit as \( x

ightarrow2^{-} \)
When \( x
ightarrow2^{-} \), \( 1 < x<2 \), so \( \llbracket x
rbracket = 1 \) and \( \llbracket -x
rbracket=\llbracket-(x)
rbracket \), since \( - 2<-x < - 1 \), \( \llbracket -x
rbracket=-2 \). Then \( f(x)=\llbracket x
rbracket+\llbracket -x
rbracket=1+(-2)=-1 \). So \( \lim_{x
ightarrow2^{-}}f(x)=-1 \).

Step3: Find right - hand limit as \( x

ightarrow2^{+} \)
When \( x
ightarrow2^{+} \), \( 2 < x<3 \), so \( \llbracket x
rbracket = 2 \) and \( \llbracket -x
rbracket=\llbracket-(x)
rbracket \), since \( - 3<-x < - 2 \), \( \llbracket -x
rbracket=-3 \). Then \( f(x)=\llbracket x
rbracket+\llbracket -x
rbracket=2+(-3)=-1 \). So \( \lim_{x
ightarrow2^{+}}f(x)=-1 \).

Step4: Find \( f(2) \)

For \( x = 2 \), \( \llbracket 2
rbracket=2 \) and \( \llbracket - 2
rbracket=-2 \). Then \( f(2)=\llbracket 2
rbracket+\llbracket - 2
rbracket=2+(-2)=0 \).

Answer:

• For \( f(a) \) (a is integer): \( 0 \)
• For \( \lim_{x

ightarrow2^{-}}f(x) \): \( - 1 \)

• For \( \lim_{x

ightarrow2^{+}}f(x) \): \( - 1 \)

• For \( \lim_{x

ightarrow2}f(x) \): \( - 1 \)

• For \( f(2)=\llbracket 2

rbracket+\llbracket - 2
rbracket=2+\underline{-2}=\underline{0} \)