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 determine the continuity of \( f(x) \) at \( x = 1 \), we need to check if the left-hand limit (LHL), the right-hand limit (RHL), and the function's value at \( x = 1 \) are all equal. 
Step 1: Calculate the Left-Hand Limit (LHL)
\[ {LHL} = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{x^2 - 1}{x^2 + 2x - 3} \] Factor the numerator and denominator: \[ \lim_{x \to 1^-} \frac{(x - 1)(x + 1)}{(x + 3)(x - 1)} \] Cancel the common factor \( (x - 1) \): \[ \lim_{x \to 1^-} \frac{x + 1}{x + 3} \] Substitute \( x = 1 \): \[ \frac{1 + 1}{1 + 3} = \frac{2}{4} = \frac{1}{2} \] 
Step 2: Calculate the Right-Hand Limit (RHL)
\[ {RHL} = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{x^2 - 1}{x^2 - 2x + 1} \] Factor the numerator and denominator: \[ \lim_{x \to 1^+} \frac{(x - 1)(x + 1)}{(x - 1)^2} \] Cancel the common factor \( (x - 1) \): \[ \lim_{x \to 1^+} \frac{x + 1}{x - 1} \] As \( x \) approaches 1 from the right (i.e., \( x>1 \)), the numerator approaches 2, and the denominator approaches 0 from the positive side. Thus, the limit is \( +\infty \). \[ \lim_{x \to 1^+} \frac{x + 1}{x - 1} = +\infty \] 
Step 3: Evaluate the function at \( x = 1 \)
From the definition of \( f(x) \), we have \( f(1) = \frac{1}{2} \). 
Step 4: Compare LHL, RHL, and \( f(1) \)
We have: \[ {LHL} = \frac{1}{2} \] \[ {RHL} = +\infty \] \[ f(1) = \frac{1}{2} \] 
Conclusion:
Since LHL \( \ne \) RHL, the limit \( \lim_{x \to 1} f(x) \) does not exist. Therefore, the function \( f(x) \) is discontinuous at \( x = 1 \). Also, although LHL = \( f(1) \), the function is still discontinuous at \( x = 1 \) because the RHL is not equal to these values. For a function to be continuous at a point, the LHL, RHL, and the value of the function at that point must all be equal. 
 

Answer 2

\[ f(x) = \frac{x^2 - 1}{x^2 - 2|x-1| - 1}. \] We first simplify the expression for \( x \neq 1 \). Notice that \( x^2 - 1 = (x - 1)(x + 1) \). Thus, the numerator becomes: \[ x^2 - 1 = (x - 1)(x + 1). \] For the denominator, consider the expression \( |x - 1| \), which behaves differently depending on whether \( x > 1 \) or \( x < 1 \). We need to evaluate the limit from both sides of \( x = 1 \).

Step 4: Evaluate the left-hand limit as \( x \to 1^- \)
For \( x \to 1^- \), \( |x - 1| = -(x - 1) \). Therefore, the denominator becomes: \[ x^2 - 2|x - 1| - 1 = x^2 + 2(x - 1) - 1 = x^2 + 2x - 3. \] Thus, for \( x \neq 1 \), we have: \[ f(x) = \frac{(x - 1)(x + 1)}{x^2 + 2x - 3}. \] This can be simplified as: \[ f(x) = \frac{(x - 1)(x + 1)}{(x - 1)(x + 3)} = \frac{x + 1}{x + 3}. \] Now, taking the limit as \( x \to 1^- \): \[ \lim_{x \to 1^-} f(x) = \frac{1 + 1}{1 + 3} = \frac{2}{4} = \frac{1}{2}. \] Step 5: Evaluate the right-hand limit as \( x \to 1^+ \)
For \( x \to 1^+ \), \( |x - 1| = x - 1 \). Therefore, the denominator becomes: \[ x^2 - 2(x - 1) - 1 = x^2 - 2x + 2 - 1 = x^2 - 2x + 1 = (x - 1)^2. \] Thus, for \( x \neq 1 \), we have: \[ f(x) = \frac{(x - 1)(x + 1)}{(x - 1)^2} = \frac{x + 1}{x - 1}. \] Taking the limit as \( x \to 1^+ \), we find that the limit does not exist, because the denominator approaches \( 0 \) and the expression tends to infinity.

Step 6: Conclusion
Since the right-hand limit does not exist, the function is not continuous at \( x = 1 \). Therefore, the correct answer is:
 

The function is not continuous at \( x = 1 \)