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:
We are calculating the limit of a function as \(x \to 1\). Direct substitution yields a \(0/0\) form.
We can resolve this by algebraic manipulation, specifically by factoring the denominator.
Step 2: Key Formula or Approach:
Recognize that \(x - 1\) is a difference of squares: \(x - 1 = (\sqrt{x} - 1)(\sqrt{x} + 1)\).
Cancel the common factor \((\sqrt{x} - 1)\) from the numerator and denominator.
Step 3: Detailed Explanation:
Write the limit expression:
\[ L = \lim_{x \to 1} \frac{(\sqrt{x + 3})(\sqrt{x} - 1)}{x - 1} \] Factor the denominator:
\[ x - 1 = (\sqrt{x})^2 - 1^2 = (\sqrt{x} - 1)(\sqrt{x} + 1) \] Substitute this back into the limit:
\[ L = \lim_{x \to 1} \frac{(\sqrt{x + 3})(\sqrt{x} - 1)}{(\sqrt{x} - 1)(\sqrt{x} + 1)} \] Cancel the \(\sqrt{x} - 1\) term (valid since \(x \to 1\), so \(\sqrt{x} - 1 \neq 0\)):
\[ L = \lim_{x \to 1} \frac{\sqrt{x + 3}}{\sqrt{x} + 1} \] Now, use direct substitution \(x = 1\):
\[ L = \frac{\sqrt{1 + 3}}{\sqrt{1} + 1} \] \[ L = \frac{\sqrt{4}}{1 + 1} = \frac{2}{2} = 1 \] Step 4: Final Answer:
The limit is 1.