Question

Check whether function \( f(x) \) defined as \( f(x) = \begin{cases \frac{|x - 3|}{2(x - 3)} & , x \\ \frac{x - 6}{6} & , x \geq 3 \end{cases} \) is continuous at \( x = 3 \) or not ?}

Hint:

Modulus functions create corners in graphs but don't always create discontinuities.
When a function involves a fraction like \( \frac{|x-a|}{x-a} \), it's essentially a jump function with values \( \pm 1 \) (or scaled versions thereof).

Answer 1

Concept:
• Continuity Check: A function \( f(x) \) is continuous at \( x = a \) if \( \text{LHL} = \text{RHL} = f(a) \).
• Modulus property: \( |x - 3| = -(x - 3) \) when \( x < 3 \).

Step 1: Calculate the Left Hand Limit (LHL) at \( x = 3 \)
For \( x < 3 \), \( |x - 3| = -(x - 3) \). \[ \text{LHL} = \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} \frac{-(x - 3)}{2(x - 3)} \] Cancelling the common term \( (x - 3) \): \[ \text{LHL} = \lim_{x \to 3^-} \left( -\frac{1}{2} \right) = -\frac{1}{2} \]

Step 2: Calculate the Right Hand Limit (RHL) at \( x = 3 \)
For \( x \geq 3 \), the function is defined as \( \frac{x - 6}{6} \). \[ \text{RHL} = \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} \frac{x - 6}{6} \] \[ \text{RHL} = \frac{3 - 6}{6} = \frac{-3}{6} = -\frac{1}{2} \]

Step 3: Compare LHL, RHL, and the function value \( f(3) \)
From the definition, \( f(3) \) is calculated using the \( x \geq 3 \) case: \[ f(3) = \frac{3 - 6}{6} = -\frac{1}{2} \] Since: \[ \text{LHL} = \text{RHL} = f(3) = -\frac{1}{2} \] The function \( f(x) \) is continuous at \( x = 3 \).