We are tasked with finding the limit of the function

$f(x) = \frac{5x^{\frac{8}{3}} - x^{\frac{7}{3}} + 7}{x^{\frac{13}{5}} + 3x + \sqrt{x}}$ as $x \to \infty$.

Step-by-Step Solution

1. Identify dominant terms: When $x \to \infty$, the highest degree terms in both the numerator and denominator will dominate, since lower-order terms become negligible in comparison to higher-order terms.

   • In the numerator, the dominant term is $5x^{\frac{8}{3}}$.
   • In the denominator, the dominant term is $x^{\frac{13}{5}}$.

2. Simplify the expression using the dominant terms: We approximate $f(x)$ for large $x$ by ignoring the lower-order terms and focusing on the dominant terms in both the numerator and denominator.

   $f(x) \sim \frac{5x^{\frac{8}{3}}}{x^{\frac{13}{5}}}$

3. Simplify the powers of $x$: Now, simplify the expression by subtracting the exponents of $x$.

   • The exponent in the numerator is $\frac{8}{3}$.
   • The exponent in the denominator is $\frac{13}{5}$.

   We need to subtract these exponents: $\frac{8}{3} - \frac{13}{5} = \frac{40}{15} - \frac{39}{15} = \frac{1}{15}$

   So we have: $f(x) \sim \frac{5}{x^{\frac{1}{15}}}$

4. Take the limit as $x \to \infty$: As $x \to \infty$, the term $x^{\frac{1}{15}}$ grows without bound, and thus the fraction $\frac{5}{x^{\frac{1}{15}}}$ approaches 0.

   Therefore, the limit is: $\lim_{x \to \infty} f(x) = 0$

Final Answer:

$\boxed{0}$

Would you like more details or have any questions?

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Here are 5 related questions you might find helpful:

1. How do you find limits when both the numerator and denominator grow without bound?
2. What are dominant terms, and why are they important in evaluating limits?
3. How do you simplify expressions involving fractional exponents?
4. How does L'Hôpital's rule apply to limits involving infinity?
5. What are some common limits involving fractional powers?

Tip: When working with limits at infinity, focus on the highest-order terms to simplify your calculation.