Question

.The function \( f(x) \) is given by: \[ f(x) = \begin{cases} \frac{2}{5 - x}, & x<3 \\ 5 - x, & x \geq 3 \end{cases} \] Which of the following is true

• A. left discontinuous at \( x = 3 \)
• B. left continuous at \( x = 3 \)
• C. right discontinuous at \( x = 5 \)
• D. discontinuous at \( x = 5 \)

Hint:

A function is left discontinuous at \( x = a \) if \( \lim_{x \to a^-} f(x) \neq f(a) \). A function is right discontinuous at \( x = a \) if \( \lim_{x \to a^+} f(x) \neq f(a) \). A function is completely discontinuous at \( x = a \) if \( LHL \neq RHL \).

Answer 1

The Correct Option is A

Step 1: Check Left-Hand and Right-Hand Limits at \( x = 3 \) To determine the continuity at \( x = 3 \), we compute the left-hand limit \( LHL \), right-hand limit \( RHL \), and function value \( f(3) \). \[ LHL = \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} \frac{2}{5 - x} \] Substituting \( x = 3 \): \[ LHL = \frac{2}{5 - 3} = \frac{2}{2} = 1 \] Now, compute the right-hand limit: \[ RHL = \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (5 - x) \] Substituting \( x = 3 \): \[ RHL = 5 - 3 = 2 \] 

Step 2: Checking Continuity at \( x = 3 \) Since \( LHL \neq RHL \), the function is discontinuous at \( x = 3 \). Since \( LHL \neq f(3) \), it is left discontinuous at \( x = 3 \), confirming option (A). 

Step 3: Checking Continuity at \( x = 5 \) To check for discontinuity at \( x = 5 \), we compute: \[ LHL = \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (5 - x) = 5 - 5 = 0 \] \[ RHL = \lim_{x \to 5^+} f(x) \] Since \( x<5 \) does not exist in the domain of the given function, there is no discontinuity at \( x = 5 \).