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The function $f(x)=|x|$ at $x=0$ is
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Verified Answer
The correct answer is:
Continuous but non-differentiable
Since this function is continuous at $x=0$
Now for differentiability
$\begin{aligned} & f(x)=|x|=|0|=0 \text { and } f(0+h)=f(h)=|h| \\ & \therefore \lim _{h \rightarrow 0-} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0-} \frac{|h|}{h}=-1 \\ & \text { and } \lim _{h \rightarrow 0+} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0+} \frac{|h|}{h}=1 .\end{aligned}$
Therefore it is continuous and non-differentiable.
Now for differentiability
$\begin{aligned} & f(x)=|x|=|0|=0 \text { and } f(0+h)=f(h)=|h| \\ & \therefore \lim _{h \rightarrow 0-} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0-} \frac{|h|}{h}=-1 \\ & \text { and } \lim _{h \rightarrow 0+} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0+} \frac{|h|}{h}=1 .\end{aligned}$
Therefore it is continuous and non-differentiable.
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