Q100 / 180 · COMEDK 2025 (10 May Shift 2) · Mathematics
The function $f(x)=\left\{\begin{array}{c}\dfrac{|x|}{x}, \text { if } x \neq 0 \\ 0, \text { if } x=0\end{array}\right.$ is discontinuous at
- Correct answerA$x=0$
- B$x <0$
- C$x>1$
- D$x>0$
Solution
The function is defined as $f(x) = \dfrac{|x|}{x}$ for $x \neq 0$ and $f(0) = 0$.
For $x > 0$, $|x| = x$, so $f(x) = \dfrac{x}{x} = 1$.
For $x < 0$, $|x| = -x$, so $f(x) = \dfrac{-x}{x} = -1$.
Evaluating the limits at $x = 0$:
Left hand limit: $\lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} (-1) = -1$.
Right hand limit: $\lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} (1) = 1$.
Since the left hand limit and right hand limit are not equal, the limit $\lim_{x \to 0} f(x)$ does not exist. Therefore, the function is discontinuous at $x = 0$.
Answer: $x=0$
For $x > 0$, $|x| = x$, so $f(x) = \dfrac{x}{x} = 1$.
For $x < 0$, $|x| = -x$, so $f(x) = \dfrac{-x}{x} = -1$.
Evaluating the limits at $x = 0$:
Left hand limit: $\lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} (-1) = -1$.
Right hand limit: $\lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} (1) = 1$.
Since the left hand limit and right hand limit are not equal, the limit $\lim_{x \to 0} f(x)$ does not exist. Therefore, the function is discontinuous at $x = 0$.
Answer: $x=0$
Hence, the correct answer is option: 'A'.