Squeeze Theorem Math Example 2

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

Example 2

hard
Use the squeeze theorem to prove limโกxโ†’0sinโกxx=1\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1.

Solution

  1. 1
    For 0<x<ฯ€20 < x < \frac{\pi}{2}, a geometric argument gives: cosโกx<sinโกxx<1\cos x < \frac{\sin x}{x} < 1.
  2. 2
    For โˆ’ฯ€2<x<0-\frac{\pi}{2} < x < 0, the inequality also holds by the symmetry of sine (odd function).
  3. 3
    limโกxโ†’0cosโกx=1\lim_{x\to 0} \cos x = 1 and limโกxโ†’01=1\lim_{x\to 0} 1 = 1.
  4. 4
    By the squeeze theorem: limโกxโ†’0sinโกxx=1\lim_{x\to 0} \frac{\sin x}{x} = 1.

Answer

limโกxโ†’0sinโกxx=1\lim_{x \to 0}\frac{\sin x}{x} = 1
This fundamental limit cannot be proved by algebra alone. The squeeze theorem traps sinโกxx\frac{\sin x}{x} between cosโกx\cos x and 1; since both bounds approach 1, so does the middle expression.

About Squeeze Theorem

If g(x)โ‰คf(x)โ‰คh(x)g(x) \leq f(x) \leq h(x) near x=ax = a, and limโกxโ†’ag(x)=limโกxโ†’ah(x)=L\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L, then limโกxโ†’af(x)=L\lim_{x \to a} f(x) = L.

Learn more about Squeeze Theorem โ†’

More Squeeze Theorem Examples

Example 1 easy

Use the squeeze theorem to find [formula].

Example 3 easy

Find [formula].

Example 4 medium

Show that [formula].

โ† All Squeeze Theorem Examples Squeeze Theorem Concept