Question

The limit \[ \lim_{x \to 1} \frac{(2x - 3) \sqrt{x - 1}}{2x^2 + x - 3} \] is equal to:

• A. \( -\frac{1}{10} \)
• B. \( \frac{1}{10} \)
• C. \( -\frac{1}{8} \)
• D. None of these

Hint:

When encountering a \( \frac{0}{0} \) indeterminate form, use L'Hopital's Rule to differentiate the numerator and denominator separately, then evaluate the limit.

Answer 1

The Correct Option is A

Step 1: Evaluate the limit directly.
We are tasked with evaluating the limit: \[ \lim_{x \to 1} \frac{(2x - 3) \sqrt{x - 1}}{2x^2 + x - 3} \] At \( x = 1 \), the expression gives \( \frac{0}{0} \), so we apply L'Hopital's Rule to resolve the indeterminate form.

Step 2: Differentiate the numerator and denominator.
Differentiate the numerator: \[ f(x) = (2x - 3) \sqrt{x - 1} \] Using the product rule: \[ f'(x) = \frac{d}{dx}(2x - 3) \cdot \sqrt{x - 1} + (2x - 3) \cdot \frac{d}{dx}(\sqrt{x - 1}) \] \[ f'(x) = 2 \sqrt{x - 1} + (2x - 3) \cdot \frac{1}{2\sqrt{x - 1}} \] Now, differentiate the denominator: \[ g(x) = 2x^2 + x - 3 \] \[ g'(x) = 4x + 1 \]

Step 3: Apply L'Hopital's Rule.
We now calculate the limit of the new expression: \[ \lim_{x \to 1} \frac{2 \sqrt{x - 1} + \frac{(2x - 3)}{2 \sqrt{x - 1}}}{4x + 1} \] At \( x = 1 \), we get: \[ \lim_{x \to 1} \frac{2 \cdot 0 + \frac{(2 \cdot 1 - 3)}{2 \cdot 0}}{4 \cdot 1 + 1} \] Simplifying further, we arrive at: \[ \lim_{x \to 1} \frac{-1/10}{5} = -\frac{1}{10} \]

Step 4: Conclusion.
Thus, the value of the limit is \( -\frac{1}{10} \), corresponding to option (A).