Using any of the results in this section, prove the following.  (d) lim (√ n /(n+1))=0 (e) lim n^2 / n! =0 (f) If Ixl < 1, then lim (n approach infinity)x^n = 0

Let's address each part of the question one by one, providing a step-by-step proof.

### Part (d): \(\lim_{n \to \infty} \frac{\sqrt{n}}{n+1} = 0\)

To evaluate the limit \(\lim_{n \to \infty} \frac{\sqrt{n}}{n+1}\), we proceed as follows:

1. **Rewrite the expression**:
   \[
   \frac{\sqrt{n}}{n+1} = \frac{\sqrt{n}}{n(1 + \frac{1}{n})}
   \]

2. **Simplify the expression**:
   \[
   \frac{\sqrt{n}}{n(1 + \frac{1}{n})} = \frac{1}{\sqrt{n}(1 + \frac{1}{n})}
   \]

3. **Analyze the limit**: As \(n \to \infty\),
   - \(\sqrt{n} \to \infty\),
   - \(\frac{1}{n} \to 0\), so \(1 + \frac{1}{n} \to 1\).

   Hence,
   \[
   \frac{1}{\sqrt{n}(1 + \frac{1}{n})} \approx \frac{1}{\sqrt{n}}
   \]

4. **Conclusion**: As \(n \to \infty\), \(\frac{1}{\sqrt{n}} \to 0\), so the original limit is
   \[
   \lim_{n \to \infty} \frac{\sqrt{n}}{n+1} = 0.
   \]

### Part (e): \(\lim_{n \to \infty} \frac{n^2}{n!} = 0\)

We need to evaluate the limit \(\lim_{n \to \infty} \frac{n^2}{n!}\).

1. **Factorial growth**: As \(n\) increases, \(n!\) (the factorial of \(n\)) grows much faster than any polynomial term like \(n^2\). Specifically, \(n! = n \cdot (n-1) \cdot (n-2) \cdots 2 \cdot 1\), which grows exponentially compared to \(n^2\).

2. **Bounding the term**: Notice that starting from \(n \geq 3\),
   \[
   \frac{n^2}{n!} = \frac{n \cdot n}{n \cdot (n-1) \cdot (n-2) \cdot \cdots \cdot 1}.
   \]
   For large \(n\), we have \(n! > n^2\).

3. **Conclusion**: Since factorial growth dominates the polynomial term, we conclude that
   \[
   \lim_{n \to \infty} \frac{n^2}{n!} = 0.
   \]

### Part (f): If \(|x| < 1\), then \(\lim_{n \to \infty} x^n = 0\)

Given \(|x| < 1\), we want to prove that \(\lim_{n \to \infty} x^n = 0\).

1. **Absolute value of \(x^n\)**: Since \(|x| < 1\), we know that
   \[
   |x^n| = |x|^n.
   \]

2. **Behavior of \(x^n\) as \(n \to \infty\)**: Since \(|x| < 1\), \(|x|^n\) decreases geometrically as \(n\) increases. Specifically, for any \(|x| < 1\), the terms of the sequence \(|x|^n\) get smaller and smaller, approaching 0 as \(n\) becomes large.

3. **Conclusion**: Since \(|x^n|\) tends to 0 as \(n \to \infty\), it follows that
   \[
   \lim_{n \to \infty} x^n = 0.
   \]

---

### Do you have any questions or would you like further details?

Here are some related questions to explore the concepts further:

1. What is the difference between the growth rates of polynomial and factorial functions?
2. How can we generalize the behavior of sequences involving powers of numbers less than 1?
3. How does the squeeze theorem help in proving limits like part (f)?
4. Can we apply L'Hopital's Rule to solve part (e)? Why or why not?
5. How do geometric series relate to the behavior of \(x^n\) when \(|x| < 1\)?

**Tip**: When dealing with limits involving factorials and powers, understanding the growth rates of functions is crucial for determining whether the limit tends to zero, infinity, or some other value.

Using any of the results in this section, prove the following. (d) lim (√n /(n+1))=0 (e) lim n^2 / n! =0 (f) If |x| < 1, then lim (n approaches infinity)x^n = 0

Explore the proof of three mathematical limits involving square roots, factorial growth, and exponential decay. This detailed explanation covers key concepts such as the asymptotic behavior of functions and the behavior of powers of numbers less than 1, suitable for college-level students.

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