Prove that (log2 n)^3 is smaller than n when n goes very large.

To prove that \((\log_2 n)^3\) is smaller than \(n\) when \(n\) goes very large, we can analyze the behavior of both functions as \(n\) approaches infinity. 

Let's denote \(f(n) = (\log_2 n)^3\) and \(g(n) = n\).

### Step 1: Comparing Growth Rates
To compare the growth rates of \(f(n)\) and \(g(n)\), we can use the concept of limits and L'Hôpital's rule. We need to show that:

\[
\lim_{n \to \infty} \frac{(\log_2 n)^3}{n} = 0
\]

### Step 2: Changing the Base of Logarithm
First, recall the change of base formula for logarithms:

\[
\log_2 n = \frac{\ln n}{\ln 2}
\]

Thus, \((\log_2 n)^3\) can be rewritten as:

\[
(\log_2 n)^3 = \left(\frac{\ln n}{\ln 2}\right)^3 = \frac{(\ln n)^3}{(\ln 2)^3}
\]

### Step 3: Applying L'Hôpital's Rule
Now we consider the limit:

\[
\lim_{n \to \infty} \frac{(\log_2 n)^3}{n} = \lim_{n \to \infty} \frac{\frac{(\ln n)^3}{(\ln 2)^3}}{n} = \frac{1}{(\ln 2)^3} \lim_{n \to \infty} \frac{(\ln n)^3}{n}
\]

Let’s focus on evaluating \(\lim_{n \to \infty} \frac{(\ln n)^3}{n}\). 

### Step 4: Applying L'Hôpital's Rule
Since both the numerator and the denominator approach infinity as \(n\) goes to infinity, we apply L'Hôpital's rule:

\[
\lim_{n \to \infty} \frac{(\ln n)^3}{n}
\]

Taking derivatives with respect to \(n\):

\[
\frac{d}{dn}[(\ln n)^3] = 3(\ln n)^2 \cdot \frac{1}{n} = \frac{3(\ln n)^2}{n}
\]

\[
\frac{d}{dn}[n] = 1
\]

Now applying L'Hôpital's rule:

\[
\lim_{n \to \infty} \frac{(\ln n)^3}{n} = \lim_{n \to \infty} \frac{\frac{3(\ln n)^2}{n}}{1} = \lim_{n \to \infty} \frac{3(\ln n)^2}{n}
\]

Applying L'Hôpital's rule again to \(\frac{(\ln n)^2}{n}\):

\[
\frac{d}{dn}[(\ln n)^2] = 2(\ln n) \cdot \frac{1}{n} = \frac{2 \ln n}{n}
\]

\[
\frac{d}{dn}[n] = 1
\]

So,

\[
\lim_{n \to \infty} \frac{3(\ln n)^2}{n} = 3 \lim_{n \to \infty} \frac{2 \ln n}{n} = 3 \cdot 2 \lim_{n \to \infty} \frac{\ln n}{n}
\]

Applying L'Hôpital's rule once more to \(\frac{\ln n}{n}\):

\[
\frac{d}{dn}[\ln n] = \frac{1}{n}
\]

\[
\frac{d}{dn}[n] = 1
\]

Thus,

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

So,

\[
\lim_{n \to \infty} \frac{(\log_2 n)^3}{n} = \frac{1}{(\ln 2)^3} \cdot 0 = 0
\]

This proves that \((\log_2 n)^3\) grows much slower than \(n\) and thus becomes negligible as \(n\) approaches infinity.

### Conclusion
\[
(\log_2 n)^3 < n \quad \text{for sufficiently large} \; n
\]

Would you like more details or have any questions? Here are five additional questions you might find interesting:

1. How does L'Hôpital's rule work for limits approaching infinity?
2. Can you explain the change of base formula for logarithms in more detail?
3. What are other common techniques to compare growth rates of functions?
4. How can we prove inequalities involving logarithms without L'Hôpital's rule?
5. What are some practical applications of understanding the growth rates of functions?

Tip: When comparing the growth rates of different functions, considering their behavior as they approach infinity can often provide clear insights into their relative magnitudes.

Learn how to prove that (log2 n)^3 is smaller than n as n becomes very large using mathematical analysis. This problem involves logarithms, limits, and L'Hôpital's rule, demonstrating advanced concepts suitable for advanced high school or undergraduate students.

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