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

Step 1: Understanding the Concept:
This limit involves the behavior of exponential growth as $x$ approaches infinity. In a sum of exponential terms, the term with the largest base dominates the growth of the entire expression.

Step 2: Detailed Explanation:
1. Let $L = \lim_{x \to \infty} \left[ \frac{1^x + 2^x + \dots + n^x}{n} \right]^{1/x}$. 2. Factor out the largest base, which is $n^x$: \[ L = \lim_{x \to \infty} \left[ \frac{n^x \left( (\frac{1}{n})^x + (\frac{2}{n})^x + \dots + (\frac{n-1}{n})^x + 1 \right)}{n} \right]^{1/x} \] 3. Distribute the exponent $1/x$: \[ L = \lim_{x \to \infty} \frac{n}{n^{1/x}} \left[ (\text{terms approaching 0}) + 1 \right]^{1/x} \] 4. As $x \to \infty$: - $n^{1/x} \to n^0 = 1$ - $(1/n)^x, (2/n)^x \dots \to 0$ - $[1]^{1/x} \to 1$ 5. The limit evaluates to: \[ L = \frac{n}{1} [1] = n \]

Step 3: Final Answer
The limit is $n$. (Note: If the limit was $x \to 0$, the answer would be the geometric mean $(n!)^{1/n}$).