The function $f(x)=|2sgn (2x)|+2$ has
(a) Jump discontinuity
(b) Removable discontinuity
(c) Infinite discontinuity
(d) No discontinuity at $x=0$

Hint: We will analyze the $sgn $ function. Then, we will look at the left limit and the right limit of $f(x)$. We will then find the value of $f(x)$ at $x=0$. The function is continuous if the left limit and the right limit are equal, and the value of the limit is equal to the value of the function at that point.

Complete step by step answer:
The $sgn $ function has the following values,
$sgn (x)=-1$ for $x<0$
$sgn (x)=0$ for $x=0$
$sgn (x)=1$ for $x>0$
Now, we will check the left limit of the given function,
$\displaystyle \lim_{x \to 0^-}f(x)=\displaystyle \lim_{x \to 0^-}|2sgn (2x)|+2$
As $x\to {{0}^{-}}$ , the value of $sgn (2x)=-1$. Therefore, we have
$\displaystyle \lim_{x \to 0^-}|2sgn (2x)|+2=|2\cdot (-1)|+2=2+2=4$
Next, we will check the right limit of the given function,
$\displaystyle \lim_{x \to 0^+}f(x)=\displaystyle \lim_{x \to 0^+}|2sgn (2x)|+2$
As $x\to {{0}^{+}}$ , the value of $sgn (2x)=1$. Therefore, we have
$\displaystyle \lim_{x \to 0^+}|2sgn (2x)|+2=|2\cdot (1)|+2=2+2=4$
So, we have that $\displaystyle \lim_{x \to 0^-}f(x)=\displaystyle \lim_{x \to 0^+}f(x)=4$. Now, we have to check the value of the function at $x=0$.
$f(0)=|2sgn (2\cdot 0)|+2=|2\cdot 0|+2=0+2=2$.
So we have $f(0)=2$.
Hence, $f(x)$ is discontinuous at $x=0$, since $\displaystyle \lim_{x \to 0^-}f(x)=\displaystyle \lim_{x \to 0^+}f(x)\ne f(0)$.
As $x=0$ is the only point of discontinuity, $f(x)$ has a removable discontinuity.

So, the correct answer is “Option B”.

Note: The definition of jump continuity requires $\displaystyle \lim_{x \to 0^-}f(x)\ne \displaystyle \lim_{x \to 0^+}f(x)$. The values of the function for all $x$ are finite, and the point of discontinuity of $f(x)$also has a finite value. Hence, it cannot be an infinite singularity. Also, since $\displaystyle \lim_{x \to 0^-}f(x)=\displaystyle \lim_{x \to 0^+}f(x)\ne f(0)$, implies that the function has a discontinuity at $x=0$. The answer for this question can also be obtained by eliminating the other options.