We are tasked with finding the limit of the function $f(x) = \frac{x^5}{\sqrt{x^{10} + 8}}$ as $x$ approaches infinity.

Step-by-Step Solution

1. Simplify the function for large $x$:

   For very large $x$, the $x^{10}$ term inside the square root will dominate the constant 8. Hence, we can approximate $\sqrt{x^{10} + 8} \approx \sqrt{x^{10}} = x^5$ for large $x$. This is because as $x \to \infty$, the constant becomes negligible compared to $x^{10}$.

   Now rewrite the function: $f(x) = \frac{x^5}{\sqrt{x^{10} + 8}} \approx \frac{x^5}{\sqrt{x^{10}}} = \frac{x^5}{x^5}$

2. Limit of the simplified expression:

   After simplifying, we get: $f(x) \approx \frac{x^5}{x^5} = 1$

   Therefore, as $x \to \infty$, the function tends toward 1.

Conclusion:

$\lim_{x \to \infty} \frac{x^5}{\sqrt{x^{10} + 8}} = 1$

Would you like further clarification or more details? Here are some related questions to deepen your understanding:

1. How do we formally justify dropping the constant term in limits?
2. What would happen if the exponent of $x$ in the numerator was different, say $x^6$?
3. How can we generalize this approach for similar functions?
4. What is the limit if we replace the constant 8 with another term involving $x$?
5. Can L'Hopital's rule be applied to this problem? Why or why not?

Tip: When dealing with limits at infinity, focusing on the dominant terms often simplifies the evaluation.