Question

Evaluate \( \lim_{x \to 1} \frac{\sqrt{x+3} \cdot \sqrt{x-1}}{x-1} \)

Hint:

When faced with indeterminate forms, use L'Hopital's Rule, which involves differentiating the numerator and denominator.

Answer 1

Step 1: Apply L'Hopital's Rule.
We observe that the expression \( \frac{\sqrt{x+3} \cdot \sqrt{x-1}}{x-1} \) results in an indeterminate form \( \frac{0}{0} \) as \( x \to 1 \). Therefore, we can apply L'Hopital's Rule, which is used to evaluate limits of indeterminate forms \( \frac{0}{0} \).
Step 2: Differentiate the numerator and denominator.
Let the numerator be \( f(x) = \sqrt{x+3} \cdot \sqrt{x-1} \) and the denominator be \( g(x) = x - 1 \). First, differentiate the numerator using the product rule: \[ f'(x) = \frac{d}{dx} \left( \sqrt{x+3} \right) \cdot \sqrt{x-1} + \sqrt{x+3} \cdot \frac{d}{dx} \left( \sqrt{x-1} \right) \] The derivative of \( \sqrt{x+3} \) is \( \frac{1}{2\sqrt{x+3}} \), and the derivative of \( \sqrt{x-1} \) is \( \frac{1}{2\sqrt{x-1}} \). The derivative of the denominator is simply \( g'(x) = 1 \).
Step 3: Evaluate the limit.
Now, substitute \( x = 1 \) into the differentiated expressions: \[ \lim_{x \to 1} \frac{f'(x)}{g'(x)} = \frac{\left( \frac{1}{2\sqrt{4}} \cdot \sqrt{0} + \sqrt{4} \cdot \frac{1}{2\sqrt{0}} \right)}{1} \] However, this expression suggests that the limit is undefined due to division by zero. Hence, we need to consider the behavior of the function more carefully, but for a clearer result, numerical or alternative methods might be applied.