To find the limit of

$\lim_{x \to \infty} \frac{5x^3 + 8}{\sqrt{36x^6 + 4}},$

we’ll follow these steps:

Step 1: Simplify the Expression

Since $x$ approaches infinity, we want to analyze the terms with the highest power of $x$ in both the numerator and the denominator, as they will dominate the behavior of the function.

1. Numerator: The highest power of $x$ in the numerator is $5x^3$.
2. Denominator: The highest power of $x$ in the denominator is under a square root. Expanding the square root term, we get $\sqrt{36x^6 + 4} \approx \sqrt{36x^6} = 6x^3$ for large $x$.

Step 2: Rewrite the Expression in Terms of the Highest Power

Now, we can rewrite the limit as:

$\lim_{x \to \infty} \frac{5x^3 + 8}{\sqrt{36x^6 + 4}} = \lim_{x \to \infty} \frac{5x^3 + 8}{6x^3}.$

Step 3: Divide Both Terms by $x^3$

To simplify further, divide both the numerator and denominator by $x^3$:

$= \lim_{x \to \infty} \frac{5 + \frac{8}{x^3}}{6}.$

Step 4: Take the Limit

As $x \to \infty$, the term $\frac{8}{x^3}$ approaches 0. So, we are left with:

$= \frac{5 + 0}{6} = \frac{5}{6}.$

Final Answer

$\lim_{x \to \infty} \frac{5x^3 + 8}{\sqrt{36x^6 + 4}} = \frac{5}{6}.$

Would you like further details on any step?

Here are some related questions:

1. How would the answer change if the numerator were $7x^3 + 10$?
2. What is the behavior of this function as $x \to -\infty$?
3. How can this problem be solved using L'Hôpital's Rule?
4. What if the denominator were $\sqrt{49x^6 + 9}$ instead?
5. How does the limit change if we add a $x^2$ term in the numerator?

Tip: In limits involving polynomials and square roots, identify the highest-power terms first to simplify calculations.