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Find all points of discontinuity of f : where f is defined by: f(x)= \begin{cases}\frac{|x|}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{cases}
CBSE | Class 12 | Continuity and Differentiability | Exercise 5.1 | Maths | NCERT
Find all points of discontinuity of f : where f is defined by: f(x)= \begin{cases}\frac{|x|}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{cases}

Solution:

The function provided is f(x)= \begin{cases}\frac{|x|}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{cases}
Also the \mathrm{f}(\mathrm{x})=|\mathrm{x}| / \mathrm{x} can be defined as,
\frac{x}{x}=1 \text { if } x>0 \text { and } \frac{-x}{x}=-1 \text { if } x<0

\Rightarrow f(x)=1 \text { if } x>0, f(x)=-1 \text { if } x<0 \text { and } f(x)=0 \text { if } x=0

Now we know that, the domain of f(x) is \mathrm{R} as f(x) is defined for x>0, x<0 and x=0. It is constant function and continuous for all x>0, f(x)=1.

It is a constant function and continuous for all x<0, f(x)=-1 .

As a result f(x) is continuous on \mathrm{R}-\{0\}.

Now,

L.H.L.= \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(-1)=-1

R.H.L.= \lim _{x \rightarrow 0-} f(x)=\lim _{x \rightarrow 0^{-}}(1)=1

Since, \lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x)

As a result, \lim _{x \rightarrow 0} f(x) does not exist and f(x) is discontinuous at only x=0.

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