Respuesta :

Recall that

[tex]\lim_{x\to b}f(x)[/tex]

exists if and only if:

[tex]\lim_{x\to b^-}f(x)=\lim_{x\to b+}f(x).[/tex]

Now, from the given graph we get that:

[tex]\begin{gathered} \lim_{x\to0^+}f(x)=2, \\ \lim_{x\to0^-}f(x)=2, \end{gathered}[/tex]

Then:

[tex]\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x).[/tex]

Therefore:

[tex]\lim_{x\to0}f(x)[/tex]

exists.

Now, from the given graph we get that:

[tex]f(0)<2.[/tex]

Therefore f(x) is not continuous at x=0.

Answer: b=0.

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