Question

The function \( f(x) = \left\{ \begin{array}{ll} \frac{|x|}{x} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{array} \right. \) is discontinuous at

• A. \( x = 0 \)
• B. \( x > 1 \)
• C. \( x > 0 \)
• D. \( x < 0 \)

Hint:

A function is continuous at a point if the limit from both sides exists and equals the value of the function at that point.

Answer 1

The Correct Option is A

Step 1: Understanding the given function.
The function is defined as:
\[ f(x) = \left\{ \begin{array}{ll} \frac{|x|}{x} & \text{if } x \neq 0 0 & \text{if } x = 0 \end{array} \right. \]
For \( x \neq 0 \), \( \frac{|x|}{x} \) is the sign function, which equals \( 1 \) for \( x > 0 \) and \( -1 \) for \( x < 0 \).

Step 2: Investigating continuity at \( x = 0 \).
We check if the function is continuous at \( x = 0 \). For continuity, the limit of \( f(x) \) as \( x \) approaches 0 must equal the value of the function at \( x = 0 \).

Step 3: Calculating the left-hand limit.
As \( x \) approaches 0 from the left (\( x \to 0^- \)), \( \frac{|x|}{x} = -1 \).
Thus:
\[ \lim_{x \to 0^-} f(x) = -1 \]

Step 4: Calculating the right-hand limit.
As \( x \) approaches 0 from the right (\( x \to 0^+ \)), \( \frac{|x|}{x} = 1 \). Thus:
\[ \lim_{x \to 0^+} f(x) = 1 \]

Step 5: Conclusion.
Since the left-hand limit and the right-hand limit are not equal, the limit does not exist at \( x = 0 \). Therefore, the function is discontinuous at \( x = 0 \). The correct answer is option (A).