Answer :
Without knowing anything else about [tex]f(x)[/tex], neither of these need be true.
Suppose
[tex]f(x)=|x-3|+7=\begin{cases}10-x&\text{for }x<3\\x+4&\text{for }x>3\\0&\text{for }x=3\end{cases}[/tex]
Then (I) isn't true because, while the limit exists as [tex]x\to3[/tex] and is equal to 7, we have [tex]f(3)=0\neq7[/tex], so [tex]f(x)[/tex] is not continuous there.
(II) also true because [tex]f(x)[/tex] is not differentiable at [tex]x=3[/tex]; that is,
[tex]\displaystyle\lim_{x\to3^-}f'(x)=\lim_{x\to3}(10-x)'=\lim_{x\to3}(-1)=-1[/tex]
but
[tex]\displaystyle\lim_{x\to3^+}f'(x)=\lim_{x\to3}(x+4)'=\lim_{x\to3}1=1[/tex]
which means the derivative does not exist at [tex]x=3[/tex].
Neither of the statements is true.
Without knowing anything else about [tex]f(x)[/tex] , neither of these needs to be true.
Suppose,
[tex]f(x)=\left | x-3 \right |+4=\begin{Bmatrix} 7-x& for\; x<3 & \\ x+1& for\; x>3 & \\ 0& for\; x=0 &\end{Bmatrix}[/tex]
Then (I) isn't true because, while the limit exists as [tex]x\rightarrow 3\\[/tex] and is equal to [tex]1[/tex], we have [tex]f(3)=0\neq 1[/tex] , so [tex]f(x)[/tex] is not continuous there.
(II) is also not true because [tex]f(x)[/tex] is not differentiable at [tex]x=3[/tex] ; that is,
[tex]\lim_{x\rightarrow 3^{-}}f'x=\lim_{x\rightarrow 3}(7-x)'=\lim_{x\rightarrow 3}(-1)=-1[/tex]
but
[tex]\lim_{x\rightarrow 3^{+}}f'x=\lim_{x\rightarrow 3}(x+1)'=\lim_{x\rightarrow 3}(1)=1[/tex]
which means the derivative does not exist at [tex]x=3[/tex] .
Learn more about limits and derivatives.
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