Let's analyze the convergence of the sequence $a_n = x^{\frac{1}{n}}$ under the two specific cases:

1. Case $x > 1$:

For $x > 1$, consider the sequence $a_n = x^{\frac{1}{n}}$.

• As $n$ increases, $\frac{1}{n}$ becomes smaller and smaller, approaching 0.
• Therefore, $x^{\frac{1}{n}}$ behaves like $x^0 = 1$ as $n$ approaches infinity.

Thus, $\lim_{n \to \infty} x^{\frac{1}{n}} = 1$.

Conclusion: The sequence $a_n = x^{\frac{1}{n}}$ converges to 1 when $x > 1$.

2. Case $0 < x < 1$:

For $0 < x < 1$, again consider the sequence $a_n = x^{\frac{1}{n}}$.

• As $n$ increases, $\frac{1}{n}$ approaches 0, just as in the previous case.
• However, since $0 < x < 1$, the sequence $x^{\frac{1}{n}}$ will approach $1$ from below.

For example, if $x = \frac{1}{2}$, then $\left(\frac{1}{2}\right)^{\frac{1}{n}}$ approaches 1 as $n$ becomes large.

Conclusion: The sequence $a_n = x^{\frac{1}{n}}$ also converges to 1 when $0 < x < 1$.

Summary:

• For both cases $x > 1$ and $0 < x < 1$, the sequence $a_n = x^{\frac{1}{n}}$ converges to 1 as $n$ approaches infinity.

Would you like any further details or have additional questions?

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Here are 8 related questions:

1. How does the sequence $a_n = \left(\frac{1}{n}\right)^n$ behave as $n$ increases?
2. What is the convergence of the sequence $a_n = e^{-\frac{n}{x}}$ for different values of $x$?
3. How does the sequence $b_n = n^{\frac{1}{n}}$ compare in terms of convergence?
4. What happens to the sequence $a_n = x^{n}$ when $0 < x < 1$?
5. How can the squeeze theorem be applied to analyze the convergence of sequences?
6. What is the convergence behavior of the sequence $a_n = n \cdot x^{\frac{1}{n}}$ for $x > 0$?
7. How does the sequence $a_n = (1 + \frac{1}{n})^n$ converge?
8. Can you find the limit of the sequence $a_n = \left(1 - \frac{1}{n}\right)^n$?

Tip: Recognizing the general behavior of exponential sequences is crucial for understanding their convergence properties.