Correct Answer - Option 2 : Discontinuous at x = 0 and has discontinuity of first kind
Concept:
f(x) is continuous at x = a, if LHL = RHL = f(a)
\(\rm\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=lim _{x \rightarrow a} f(x)\)
f(x) is differentiable if LHD = RHD
\(\begin{array}{l} \rm L H D=\lim _{h \rightarrow 0^{-}} \frac{f(a-h)-f(a)}{-h} \\ \rm R H D=\lim _{h \rightarrow 0^{+}} \frac{f(a+h)-f(a)}{h} \end{array}\)
Discontinuity of the First Kind: A function f(x) is said to have a discontinuity of the first kind from the right at x = a if the right hand of the function exists but not equal to f(a).
Discontinuity of the Second Kind: A function f(x) is said to have discontinuity of the second kind at x = a, if neither left-hand limit of f(x) at x = a nor right-hand limit of f(x) at x = a exists.
Removable Discontinuity: A function f(x) is said to have a removable discontinuity at x = a if the left-hand limit at x tends to point ‘a’ is equal to the right-hand limit at x tends to point ‘a’ but their common value is not equal to f(a).
Calculation:
\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{\left| x \right|}}{x}}\\ {0,} \end{array}\begin{array}{*{20}{c}} {,x ≠ 0}\\ {x = 0} \end{array}} \right.\)
For x ≠ 0,
f(x) = -x/x = -1, if x < 0
f(x) = x/x = 1. if x > 0
Now,
LHL = \(\mathop {\lim }\limits_{x \to 0^- }f(x)=\mathop {\lim }\limits_{x \to 0^- }-x/x=-1\)
RHL = \(\mathop {\lim }\limits_{x \to 0^+ }f(x)=\mathop {\lim }\limits_{x \to 0^+}x/x=1\)
\(\mathop {\lim }\limits_{x \to 0 }f(x) = 0\)
Since,
LHL ≠ RHL we can say that the function is not continuous at x = 0
Only x = 0, is the point of discontinuity.
Based on the definition, the function has discontinuity of the first kind.
