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Limits, Sequences, and the Squeeze Theorem

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    Name
    Vu Hung
    Twitter

Problem Statement

In Calculus, evaluating the limit of a function as xx \to \infty is usually straightforward. However, when dealing with discrete Sequences, or functions that oscillate wildly, direct limit evaluation often fails.

To solve this, we bridge the gap between inequalities and calculus using the Squeeze Theorem (also known as the Sandwich Theorem).

The theorem states that if you have three functions (or sequences) f(n)f(n), g(n)g(n), and h(n)h(n) such that: f(n)g(n)h(n) for all sufficiently large nf(n) \le g(n) \le h(n) \text{ for all sufficiently large } n And you know that the outer functions converge to the same limit LL: limnf(n)=Landlimnh(n)=L\lim_{n \to \infty} f(n) = L \quad \text{and} \quad \lim_{n \to \infty} h(n) = L Then the middle function g(n)g(n) is "squeezed" between them, and must also converge to LL: limng(n)=L\lim_{n \to \infty} g(n) = L

Consider the sequence defined by: un=sin(n)nu_n = \frac{\sin(n)}{n}

We want to find the limit of this sequence as nn approaches infinity: limnsin(n)n\lim_{n \to \infty} \frac{\sin(n)}{n}.

(a) Write down an algebraic inequality that bounds the trigonometric function sin(n)\sin(n) for all real numbers nn.

(b) Divide this inequality by nn (assuming n>0n > 0) to create a bounded inequality for the sequence unu_n.

(c) Evaluate the limits of the lower bound and upper bound as nn \to \infty.

(d) Apply the Squeeze Theorem to determine limnun\lim_{n \to \infty} u_n.


Hints

  • Part (a): What is the absolute maximum value the sine curve can reach? What is the absolute minimum?
  • Part (b): Take your inequality from (a), e.g., Asin(n)BA \le \sin(n) \le B, and divide all three parts by nn. Because nn \to \infty, we assume nn is positive, so the inequality signs do not flip.
  • Part (c): You are evaluating limnAn\lim_{n \to \infty} \frac{A}{n} and limnBn\lim_{n \to \infty} \frac{B}{n}. What happens to a fraction when the denominator grows infinitely large while the numerator stays constant?
  • Part (d): If the lower and upper bounds approach the same value, state the theorem and conclude the final limit.

Solutions

Part (a): Bounding the Sine Function

  1. The standard sine function, sin(x)\sin(x), oscillates endlessly between a peak of 1 and a trough of -1.
  2. Therefore, for any real number nn: 1sin(n)1-1 \le \sin(n) \le 1

Part (b): Creating the Sequence Bounds

  1. We need to construct the sequence un=sin(n)nu_n = \frac{\sin(n)}{n} in the middle of our inequality.
  2. Divide all three parts of the inequality by nn. Since we are investigating the limit as nn \to \infty, we assume nn is a positive integer (n>0n > 0), meaning the inequality signs remain unchanged: 1nsin(n)n1n\frac{-1}{n} \le \frac{\sin(n)}{n} \le \frac{1}{n}

Part (c): Evaluating the Outer Limits

  1. Evaluate the limit of the lower bound function f(n)=1nf(n) = \frac{-1}{n}: limn1n=0\lim_{n \to \infty} \frac{-1}{n} = 0 (A constant divided by an infinitely large number approaches zero).
  2. Evaluate the limit of the upper bound function h(n)=1nh(n) = \frac{1}{n}: limn1n=0\lim_{n \to \infty} \frac{1}{n} = 0

Part (d): Applying the Squeeze Theorem

  1. We have established that for n>0n > 0: 1nsin(n)n1n\frac{-1}{n} \le \frac{\sin(n)}{n} \le \frac{1}{n}
  2. We have also proven that: limn1n=0andlimn1n=0\lim_{n \to \infty} \frac{-1}{n} = 0 \quad \text{and} \quad \lim_{n \to \infty} \frac{1}{n} = 0
  3. Because the sequence sin(n)n\frac{\sin(n)}{n} is "squeezed" between two sequences that both converge to 0, it has nowhere else to go.
  4. By the Squeeze Theorem: limnsin(n)n=0\lim_{n \to \infty} \frac{\sin(n)}{n} = 0

Takeaways

  • Taming Oscillation: The Squeeze Theorem is the definitive tool for evaluating limits of functions containing oscillating terms like sin(x)\sin(x) or cos(x)\cos(x), or alternating sequences involving (1)n(-1)^n.
  • The Art of Bounding: The hardest part of the Squeeze Theorem is usually inventing the upper and lower bounds. Look for terms in the numerator that you can replace with constants (like turning sin(x)\sin(x) into 11).
  • Connecting the Syllabus: This theorem perfectly bridges the algebraic inequality techniques of Extension 2 with the limit and calculus concepts introduced in Advanced Mathematics.

Further Readings


Connect with me

Evaluating complex limits is a crucial skill for university-level calculus. To practice more Squeeze Theorem problems, check out the booklets on Vu's Maths Hub. I post detailed walkthroughs of challenging sequence problems on my YouTube channel. Follow me on Instagram and subscribe to my Substack to learn how limits form the basis of all modern physics!