Since 
lim 
x → ∞ 
ln(x)
8x
 = 0,
 then we can conclude the following.
lim 
x → ∞ 
(8x − ln(x))	 = 	
lim 
x → ∞ 

8x 

1 − 
ln(x)
8x


 
 = 	∞ · (1 − 0) 

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?

---

**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.

Since lim x → ∞ ln(x)/8x = 0, then we can conclude the following. lim x → ∞ (8x − ln(x)) = lim x → ∞ 8x(1 − ln(x)/8x) = ∞ · (1 − 0)

Explore the limit of (8x - ln(x)) as x approaches infinity with a detailed solution, highlighting how polynomial terms outpace logarithmic growth. This calculus problem demonstrates using limits at infinity to determine the behavior of functions. Suitable for high school calculus students in Grades 11-12.

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