Question

$$ \lim_{x \to \infty} \left[ \frac{1^x + 2^x + 3^x + \cdots + n^x}{n} \right]^{1/x} = $$

• A. $(n!) n$
• B. $(n!)^{1/n}$
• C. $n$
• D. $\ln(n!)$

Hint:

For limits of the form $\lim_{x \to \infty} (\sum a_i^x)^{1/x}$, the result is always $\max(a_1, a_2, \dots, a_n)$. The largest base "wins" because it grows infinitely faster than the others.

Answer 1

The Correct Option is C

To solve the problem \(\lim_{x \to \infty} \left[ \frac{1^x + 2^x + 3^x + \cdots + n^x}{n} \right]^{1/x}\), we need to understand the behavior of the expression within the limit as \(x\) approaches infinity.

1. Let's consider the expression inside the limit: \(\frac{1^x + 2^x + 3^x + \cdots + n^x}{n}\).
2. Notice that as \(x\) becomes very large, the term \(n^x\) becomes significantly larger than the other terms \(1^x, 2^x, \dots, (n-1)^x\) because \(n^x\) grows exponentially faster than any other \(k^x\) for \(k < n\).
3. This dominance of \(n^x\) means that: \(1^x + 2^x + \cdots + n^x \approx n^x\)for large \(x\).
4. The expression simplifies to: \(\frac{n^x}{n} = n^{x-1}\).
5. Now consider the entire expression inside the limit raised to the power of \(1/x\): \(\left(n^{x-1}\right)^{1/x} = n^{(x-1)/x}\).
6. As \(x \to \infty\), the exponent \(\frac{x-1}{x} \to 1\) because: \(\frac{x-1}{x} = 1 - \frac{1}{x}\) and \(\frac{1}{x} \to 0\).
7. Thus, the limit becomes: \(\lim_{x \to \infty} n^{(x-1)/x} = n^1 = n\).

Therefore, the value of the limit is \(n\). This means the correct option is \(n\). Other options like \((n!) n\), \((n!)^{1/n}\), or \(\ln(n!)\) do not match the behavior of the limit as demonstrated.