The given limit is:

$\lim_{x \to \infty} \frac{\sqrt{9x^2+1}}{\sqrt{4x^2+1}}$

Step 1: Check for Indeterminate Form

For large $x$, the dominant term in both the numerator and denominator is the term with $x^2$. If we substitute $x \to \infty$, we get:

$\frac{\sqrt{9x^2+1}}{\sqrt{4x^2+1}} \approx \frac{\sqrt{9x^2}}{\sqrt{4x^2}} = \frac{3x}{2x} = \frac{3}{2}$

Since we have a $\frac{\infty}{\infty}$ form, we can apply L'Hôpital's Rule.

Step 2: Differentiate the Numerator and Denominator

We use the derivative of $\sqrt{f(x)}$, which is:

$\frac{d}{dx} \sqrt{f(x)} = \frac{f'(x)}{2\sqrt{f(x)}}$

Differentiating the numerator:

$\frac{d}{dx} \sqrt{9x^2+1} = \frac{18x}{2\sqrt{9x^2+1}} = \frac{18x}{2\sqrt{9x^2+1}} = \frac{9x}{\sqrt{9x^2+1}}$

Differentiating the denominator:

$\frac{d}{dx} \sqrt{4x^2+1} = \frac{8x}{2\sqrt{4x^2+1}} = \frac{4x}{\sqrt{4x^2+1}}$

Step 3: Compute the New Limit

Now, we take the limit:

$\lim_{x \to \infty} \frac{9x/\sqrt{9x^2+1}}{4x/\sqrt{4x^2+1}}$

Rewriting:

$\lim_{x \to \infty} \frac{9x}{\sqrt{9x^2+1}} \times \frac{\sqrt{4x^2+1}}{4x}$

For large $x$:

$\frac{9x}{\sqrt{9x^2+1}} \approx \frac{9x}{\sqrt{9x^2}} = \frac{9x}{3x} = 3$

$\frac{\sqrt{4x^2+1}}{4x} \approx \frac{\sqrt{4x^2}}{4x} = \frac{2x}{4x} = \frac{1}{2}$

Thus, the limit simplifies to:

$3 \times \frac{1}{2} = \frac{3}{2}$

Final Answer:

$\lim_{x \to \infty} \frac{\sqrt{9x^2+1}}{\sqrt{4x^2+1}} = \frac{3}{2}$

Would you like a more detailed breakdown or another method?

Here are five related questions:

1. How do we apply L'Hôpital's Rule to other indeterminate forms like $0/0$?
2. What are alternative methods for solving this limit, such as factoring?
3. How does the asymptotic approximation help in limit calculations?
4. When does L'Hôpital's Rule fail to give the correct limit?
5. How can we use Taylor series to approximate square root functions in limits?

Tip: Before using L'Hôpital's Rule, always check if algebraic simplifications can make the problem easier.