To analyze the limit

$\lim_{x \to \infty} (8x - \ln(x)),$

we'll use the given information:

$\lim_{x \to \infty} \frac{\ln(x)}{8x} = 0.$

This implies that $\ln(x)$ grows much slower than $8x$ as $x \to \infty$.

Now, let's proceed with the limit step-by-step:

1. Rewrite the Expression
   Start by rewriting the expression as follows:

   $\lim_{x \to \infty} \left( 8x - \ln(x) \right) = \lim_{x \to \infty} \left( 8x \cdot \left(1 - \frac{\ln(x)}{8x} \right) \right).$

2. Substitute the Limit Result
   Since $\lim_{x \to \infty} \frac{\ln(x)}{8x} = 0$, we can substitute this into the expression:

   $= \lim_{x \to \infty} 8x \cdot \left(1 - 0\right) = \lim_{x \to \infty} 8x \cdot 1.$

3. Evaluate the Limit
   Now we have:

   $\lim_{x \to \infty} 8x = \infty.$

Thus, we conclude that:

$\lim_{x \to \infty} (8x - \ln(x)) = \infty.$

This makes sense intuitively as well, since $8x$ grows unboundedly faster than $\ln(x)$ as $x$ becomes large.

Would you like a deeper breakdown or have any questions?

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Follow-up Questions:

1. How does the rate of growth of $8x$ compare to $\ln(x)$ as $x \to \infty$?
2. What does the result $\lim_{x \to \infty} \frac{\ln(x)}{8x} = 0$ imply about their growth rates?
3. How would the limit change if we replaced $8x$ with $x^2$?
4. Can we apply L'Hôpital's Rule to solve similar limits involving $\ln(x)$ and polynomial terms?
5. How does the presence of $\ln(x)$ affect limits with other higher power terms?

Tip: When comparing growth rates, logarithmic functions increase very slowly relative to polynomial functions or exponential functions, which can guide you in solving limits at infinity.