Question

Evaluate the limit \( \lim_{x \to 1} \frac{x^2 - 1}{\sqrt{x} - 1} \)

Hint:

When faced with an indeterminate form \( \frac{0}{0} \), apply L'Hopital's Rule by differentiating the numerator and denominator.

Answer 1

Correct Answer: 4

Step 1: Factor the numerator.
The expression can be factored as: \[ x^2 - 1 = (x - 1)(x + 1) \] Thus, the limit becomes: \[ \lim_{x \to 1} \frac{(x - 1)(x + 1)}{\sqrt{x} - 1} \] Step 2: Apply L'Hopital's Rule.
The limit results in the indeterminate form \( \frac{0}{0} \). Therefore, we apply L'Hopital's Rule, which involves taking the derivative of the numerator and denominator separately.
The derivative of the numerator \( (x - 1)(x + 1) \) is: \[ \frac{d}{dx}[(x - 1)(x + 1)] = 2x \] The derivative of the denominator \( \sqrt{x} - 1 \) is: \[ \frac{d}{dx}[\sqrt{x} - 1] = \frac{1}{2\sqrt{x}} \] Step 3: Calculate the limit.
Now, substitute into the limit: \[ \lim_{x \to 1} \frac{2x}{\frac{1}{2\sqrt{x}}} = \lim_{x \to 1} 4x\sqrt{x} = 4 \times 1 \times 1 = 2 \] Step 4: Conclusion.
Thus, the value of the limit is \( \boxed{2} \).