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The given limit is:

\[
\lim_{n \to \infty} \frac{(\ln n)^2}{8n}
\]

### Solution:

We are tasked with finding the limit of the function \(\frac{(\ln n)^2}{8n}\) as \(n\) approaches infinity.

1. The numerator, \((\ln n)^2\), grows slowly as \(n \to \infty\) because the natural logarithm function grows slower than any polynomial or exponential function.

2. The denominator, \(8n\), grows linearly, which is much faster than the logarithmic growth of \((\ln n)^2\).

To confirm this, we can use **L'Hopital's Rule**, which is applied to limits of the form \(\frac{f(n)}{g(n)}\) where both \(f(n)\) and \(g(n)\) approach either 0 or \(\infty\). Here both the numerator and denominator go to infinity as \(n \to \infty\), so we apply L'Hopital's Rule by differentiating the numerator and denominator.

#### Applying L'Hopital's Rule:

We first differentiate the numerator and the denominator with respect to \(n\):

- Derivative of \((\ln n)^2\):
  \[
  \frac{d}{dn} \left( (\ln n)^2 \right) = 2 \ln n \cdot \frac{1}{n}
  \]

- Derivative of \(8n\):
  \[
  \frac{d}{dn} (8n) = 8
  \]

Thus, applying L'Hopital's Rule gives us the new limit:

\[
\lim_{n \to \infty} \frac{2 \ln n \cdot \frac{1}{n}}{8} = \lim_{n \to \infty} \frac{\ln n}{4n}
\]

We apply L'Hopital's Rule again because this is still an indeterminate form \(\frac{\infty}{\infty}\):

- Derivative of \(\ln n\):
  \[
  \frac{d}{dn} \ln n = \frac{1}{n}
  \]

- Derivative of \(4n\):
  \[
  \frac{d}{dn} (4n) = 4
  \]

So we get:

\[
\lim_{n \to \infty} \frac{\frac{1}{n}}{4} = \lim_{n \to \infty} \frac{1}{4n} = 0
\]

Thus, the original limit is:

\[
\boxed{0}
\]

### Conclusion:
The limit of \(\frac{(\ln n)^2}{8n}\) as \(n \to \infty\) is \(0\).

Would you like more details or have any questions?

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### Follow-up Questions:
1. How does L'Hopital's Rule help in finding limits involving infinity?
2. What other techniques can be used to solve limits like this one?
3. Can we compare the growth rates of logarithmic, linear, and exponential functions?
4. How does the behavior of logarithmic functions differ from polynomial functions at infinity?
5. What are some common scenarios where L'Hopital's Rule is applicable?

**Tip**: L'Hopital's Rule is particularly helpful when dealing with indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Keep practicing it to become proficient with complex limits.

Find the limit of (ln(n))^2 / 8n as n approaches infinity.

Explore how to find the limit of (ln(n))^2 / 8n as n approaches infinity using L'Hopital's Rule. This step-by-step solution explains logarithmic growth and applying differentiation to indeterminate forms. Suitable for university-level calculus students.

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