Question

Let $f(x) = \frac{(\sqrt{x}+3)(\sqrt{x}-1)}{x - 1}$ for $x \neq 1$. Then $\lim_{x \to 1} f(x)$ is equal to

• A. 2
• B. 4
• C. $\sqrt{2}$
• D. 1

Hint:

The difference of squares $a^2 - b^2 = (a-b)(a+b)$ is extremely common in limit problems involving square roots. Recognize $x-1$ as $(\sqrt{x}-1)(\sqrt{x}+1)$ to avoid complex differentiation.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Direct substitution results in a $0/0$ indeterminate form. We can solve this by factoring the denominator or by using L'Hopital's Rule.

Step 2: Detailed Explanation:
1. Note that $x - 1 = (\sqrt{x})^2 - 1^2 = (\sqrt{x} - 1)(\sqrt{x} + 1)$.
2. Substitute this into the expression for $f(x)$:
\[ \lim_{x \to 1} \frac{(\sqrt{x} + 3)(\sqrt{x} - 1)}{(\sqrt{x} - 1)(\sqrt{x} + 1)} \]
3. Cancel the common factor $(\sqrt{x} - 1)$:
\[ \lim_{x \to 1} \frac{\sqrt{x} + 3}{\sqrt{x} + 1} \]
4. Apply direct substitution $x = 1$:
\[ \frac{\sqrt{1} + 3}{\sqrt{1} + 1} = \frac{1 + 3}{1 + 1} = \frac{4}{2} = 2 \]

Step 3: Final Answer:
The value of the limit is 2.