Squeeze Theorem Math Example 4

Follow the full solution, then compare it with the other examples linked below.

Example 4

medium
Show that lim⁡n→∞sin⁡nn=0\displaystyle\lim_{n\to\infty} \frac{\sin n}{n} = 0.

Solution

  1. 1
    −1≤sin⁡n≤1-1 \leq \sin n \leq 1 for all integers nn.
  2. 2
    Dividing by n>0n > 0: −1n≤sin⁡nn≤1n-\frac{1}{n} \leq \frac{\sin n}{n} \leq \frac{1}{n}.
  3. 3
    lim⁡n→∞1n=0\lim_{n\to\infty} \frac{1}{n} = 0, so by the squeeze theorem, lim⁡n→∞sin⁡nn=0\lim_{n\to\infty}\frac{\sin n}{n} = 0.

Answer

00
The squeeze theorem works for sequences too. Bounding sin⁡n\sin n between -1 and 1, then dividing by a growing nn, forces the middle term to 0.

About Squeeze Theorem

If g(x)≤f(x)≤h(x)g(x) \leq f(x) \leq h(x) near x=ax = a, and lim⁡x→ag(x)=lim⁡x→ah(x)=L\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L, then lim⁡x→af(x)=L\lim_{x \to a} f(x) = L.

Learn more about Squeeze Theorem →

More Squeeze Theorem Examples

Example 1 easy

Use the squeeze theorem to find [formula].

Example 2 hard

Use the squeeze theorem to prove [formula].

Example 3 easy

Find [formula].

← All Squeeze Theorem Examples Squeeze Theorem Concept