what is the limit as x approaches infinity f(x)= (x^5)/squareroot((x^10)+8)

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.

What is the limit as x approaches infinity for f(x) = (x^5) / sqrt((x^10) + 8)?

Find the limit of the function (x^5) / sqrt((x^10) + 8) as x approaches infinity with a detailed step-by-step solution. Learn how to simplify expressions involving dominant terms at infinity. Suitable for students in Grades 11-12 studying limits and asymptotic behavior.

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