Question

Let \( f \) be the function defined by:
\[ f(x) = \begin{cases} \frac{x^2 - 1}{x^2 - 2|x-1| - 1}, & \text{if } x \neq 1, \\ \frac{1}{2}, & \text{if } x = 1. \end{cases} \] The function is continuous at:

• A. The function is continuous for all values of \( x \)
• B. The function is continuous only for \( x>1 \)
• C. The function is continuous at \( x = 1 \)
• D. The function is not continuous at \( x = 1 \)

Hint:

For a function to be continuous at a point, the left-hand limit, right-hand limit, and the function's value at that point must all exist and be equal.

Answer 1

The Correct Option is D

To assess the continuity of \( f(x) \) at \( x = 1 \), we must verify that the left-hand limit (LHL), the right-hand limit (RHL), and the function's value at \( x = 1 \) are congruent.
Step 1: Compute the Left-Hand Limit (LHL)
\[ {LHL} = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{x^2 - 1}{x^2 + 2x - 3} \] By factoring the numerator and denominator: \[ \lim_{x \to 1^-} \frac{(x - 1)(x + 1)}{(x + 3)(x - 1)} \] Upon canceling the common factor \( (x - 1) \): \[ \lim_{x \to 1^-} \frac{x + 1}{x + 3} \] Substituting \( x = 1 \): \[ \frac{1 + 1}{1 + 3} = \frac{2}{4} = \frac{1}{2} \] 
Step 2: Compute the Right-Hand Limit (RHL)
\[ {RHL} = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{x^2 - 1}{x^2 - 2x + 1} \] Factoring the numerator and denominator yields: \[ \lim_{x \to 1^+} \frac{(x - 1)(x + 1)}{(x - 1)^2} \] Eliminating the common factor \( (x - 1) \): \[ \lim_{x \to 1^+} \frac{x + 1}{x - 1} \] As \( x \) approaches 1 from the right (where \( x > 1 \)), the numerator approaches 2, and the denominator approaches 0 from the positive side. Consequently, the limit is \( +\infty \). \[ \lim_{x \to 1^+} \frac{x + 1}{x - 1} = +\infty \] 
Step 3: Determine the function's value at \( x = 1 \)
According to the definition of \( f(x) \), \( f(1) = \frac{1}{2} \). 
Step 4: Compare LHL, RHL, and \( f(1) \)
The computed values are: \[ {LHL} = \frac{1}{2} \] \[ {RHL} = +\infty \] \[ f(1) = \frac{1}{2} \] 
Conclusion:
Since LHL \( ≠ \) RHL, the limit \( \lim_{x \to 1} f(x) \) is undefined. Therefore, \( f(x) \) is discontinuous at \( x = 1 \). Although LHL = \( f(1) \), the function remains discontinuous at \( x = 1 \) due to the inequality of the RHL with these values. For a function to be continuous at a specific point, the LHL, RHL, and the function's value at that point must all be identical.