To solve the limit
[tex]\[
\lim_{x \rightarrow \infty}\left(\frac{1^{1/x} + 2^{1/x} + 3^{1/x} + \ldots + n^{1/x}}{n} \right)^{nx},
\][/tex]
we need to understand the behavior of the expression inside the limit as [tex]\(x\)[/tex] approaches infinity.
Consider the expression inside the limit,
[tex]\[
\frac{1^{1/x} + 2^{1/x} + 3^{1/x} + \ldots + n^{1/x}}{n}.
\][/tex]
For large values of [tex]\(x\)[/tex], [tex]\(i^{1/x}\)[/tex] approaches 1 for any [tex]\(i\)[/tex] because raising any positive number to the power of [tex]\(0\)[/tex] (since [tex]\(\frac{1}{x} \to 0\)[/tex] as [tex]\(x \to \infty\)[/tex]) approaches 1. Specifically,
[tex]\[
\lim_{x \to \infty} i^{1/x} = 1.
\][/tex]
This suggests that each term [tex]\(i^{1/x}\)[/tex] in the sum approaches 1 as [tex]\(x\)[/tex] goes to infinity. Consequently, the average of these [tex]\(n\)[/tex] terms also approaches 1:
[tex]\[
\lim_{x \to \infty} \frac{1^{1/x} + 2^{1/x} + \ldots + n^{1/x}}{n} = \frac{1 + 1 + \ldots + 1 (n \text{ times})}{n} = 1.
\][/tex]
Now consider the entire expression:
[tex]\[
\left( \frac{1^{1/x} + 2^{1/x} + \ldots + n^{1/x}}{n} \right)^{nx}.
\][/tex]
Since the inner fraction approaches 1 as [tex]\(x \to \infty\)[/tex], we examine the behavior of [tex]\(1^{nx}\)[/tex]:
[tex]\[
\left(1\right)^{nx} = 1.
\][/tex]
So intuitively, this expression is:
[tex]\[
\left( 1 \right)^{nx}.
\][/tex]
As [tex]\(1^y = 1\)[/tex] for any real [tex]\(y\)[/tex], it follows that
[tex]\[
\lim_{x \rightarrow \infty}\left(\frac{1^{1/x} + 2^{1/x} + 3^{1/x} + \ldots + n^{1/x}}{n} \right)^{nx} = 1.
\][/tex]
Therefore, the limit is:
[tex]\[
\boxed{1}.
\][/tex]
