Math Problem Statement

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

Solution

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

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

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

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

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

  3. Analyze the limit: As nn \to \infty,

    • n\sqrt{n} \to \infty,
    • 1n0\frac{1}{n} \to 0, so 1+1n11 + \frac{1}{n} \to 1.

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

  4. Conclusion: As nn \to \infty, 1n0\frac{1}{\sqrt{n}} \to 0, so the original limit is limnnn+1=0.\lim_{n \to \infty} \frac{\sqrt{n}}{n+1} = 0.

Part (e): limnn2n!=0\lim_{n \to \infty} \frac{n^2}{n!} = 0

We need to evaluate the limit limnn2n!\lim_{n \to \infty} \frac{n^2}{n!}.

  1. Factorial growth: As nn increases, n!n! (the factorial of nn) grows much faster than any polynomial term like n2n^2. Specifically, n!=n(n1)(n2)21n! = n \cdot (n-1) \cdot (n-2) \cdots 2 \cdot 1, which grows exponentially compared to n2n^2.

  2. Bounding the term: Notice that starting from n3n \geq 3, n2n!=nnn(n1)(n2)1.\frac{n^2}{n!} = \frac{n \cdot n}{n \cdot (n-1) \cdot (n-2) \cdot \cdots \cdot 1}. For large nn, we have n!>n2n! > n^2.

  3. Conclusion: Since factorial growth dominates the polynomial term, we conclude that limnn2n!=0.\lim_{n \to \infty} \frac{n^2}{n!} = 0.

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

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

  1. Absolute value of xnx^n: Since x<1|x| < 1, we know that xn=xn.|x^n| = |x|^n.

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

  3. Conclusion: Since xn|x^n| tends to 0 as nn \to \infty, it follows that limnxn=0.\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 xnx^n when x<1|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.

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Math Problem Analysis

Mathematical Concepts

Limits
Asymptotic Behavior
Factorial Growth
Sequences and Series

Formulas

lim_{n \to \infty} \frac{\sqrt{n}}{n+1} = 0
lim_{n \to \infty} \frac{n^2}{n!} = 0
lim_{n \to \infty} x^n = 0 (when |x| < 1)

Theorems

Limit of a Function
Factorial Growth Rate
Geometric Series
Behavior of Powers of Numbers Less Than 1

Suitable Grade Level

College Level