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The problem is asking to determine the limit:

\[
\lim_{{x \to 1}} \frac{x^2 - 1}{\sqrt{x} + 3 - \sqrt{4x}}
\]

We can proceed by solving the limit step by step:

### Step 1: Substituting \( x = 1 \)
First, substitute \( x = 1 \) directly into the function to check if we get an indeterminate form:

\[
\frac{1^2 - 1}{\sqrt{1} + 3 - \sqrt{4(1)}} = \frac{1 - 1}{1 + 3 - 2} = \frac{0}{2} = 0
\]

Since the result is not indeterminate and evaluates to \( 0 \), the limit is simply 0.

Thus:

\[
\lim_{{x \to 1}} \frac{x^2 - 1}{\sqrt{x} + 3 - \sqrt{4x}} = 0
\]

Would you like further clarification or details?

Here are some follow-up questions to deepen your understanding:

1. What happens if direct substitution in a limit leads to an indeterminate form like \( \frac{0}{0} \)?
2. Can this limit be solved using L'Hôpital's Rule, and why or why not?
3. How would you approach the limit if the expression was more complex, involving cubes or higher powers?
4. What are some alternative techniques to evaluate limits aside from direct substitution?
5. How does the presence of square roots in limits affect the strategy for solving them?

**Tip:** Always check if direct substitution works before trying more advanced techniques like factoring or L'Hôpital's Rule.

Tentukan nilai limit berikut ini (jika ada).
(a) \( \lim_{{x \to 1}} \frac{x^2 - 1}{\sqrt{x} + 3 - \sqrt{4x}} \)

Learn how to calculate the limit of the expression \( \lim_{{x \to 1}} \frac{x^2 - 1}{\sqrt{x} + 3 - \sqrt{4x}} \) using the direct substitution method. Understand key calculus concepts, including how to handle square roots in limits and when to apply L'Hôpital's Rule. Suitable for Grades 11-12.

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