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To solve the limit

\[
\lim_{{x \to \infty}} \frac{\sqrt{9x + 1}}{\sqrt{x + 1}}
\]

we can analyze the behavior of the function as \( x \to \infty \). Here's the solution step-by-step:

### Step 1: Factor the Leading Term
Rewrite the square roots to highlight the leading terms of \( x \) in both the numerator and the denominator:

\[
\frac{\sqrt{9x + 1}}{\sqrt{x + 1}} = \frac{\sqrt{x(9 + \frac{1}{x})}}{\sqrt{x(1 + \frac{1}{x})}}
\]

### Step 2: Simplify by Extracting \( \sqrt{x} \)
Factor out \( \sqrt{x} \) from both the numerator and the denominator:

\[
= \frac{\sqrt{x} \cdot \sqrt{9 + \frac{1}{x}}}{\sqrt{x} \cdot \sqrt{1 + \frac{1}{x}}}
\]

Now, we can cancel \( \sqrt{x} \) from the numerator and the denominator:

\[
= \frac{\sqrt{9 + \frac{1}{x}}}{\sqrt{1 + \frac{1}{x}}}
\]

### Step 3: Take the Limit as \( x \to \infty \)
As \( x \to \infty \), \( \frac{1}{x} \) approaches 0. So the expression inside the square roots simplifies:

\[
= \frac{\sqrt{9 + 0}}{\sqrt{1 + 0}} = \frac{\sqrt{9}}{\sqrt{1}} = \frac{3}{1} = 3
\]

### Final Answer
\[
\lim_{{x \to \infty}} \frac{\sqrt{9x + 1}}{\sqrt{x + 1}} = 3
\]

Would you like further details on any of these steps or have additional questions?

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**Related Questions:**
1. What is the limit of \(\lim_{{x \to \infty}} \frac{\sqrt{4x + 1}}{\sqrt{x + 2}}\)?
2. How does factoring out the dominant term help in solving limits at infinity?
3. Can this method be used for limits involving other roots, like cube roots?
4. What are other methods for solving limits involving square roots?
5. How do we handle limits approaching infinity in trigonometric functions?

**Tip:** When working with square roots in limits, factoring out the largest power can simplify the problem.

Evaluate the limit \( \lim_{{x \to \infty}} \frac{\sqrt{9x + 1}}{\sqrt{x + 1}} \).

Learn how to calculate \( \lim_{{x \to \infty}} \frac{\sqrt{9x + 1}}{\sqrt{x + 1}} \) using simplification techniques and the limit properties. This problem covers key calculus concepts related to limits, asymptotic behavior, and handling square roots. Suitable for students in Grades 11-12.

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