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Limits & Continuity · Continuity & Theorems

Squeeze Theorem

g(x)≤f(x)≤h(x) and lim⁡x→ag(x)=lim⁡x→ah(x)=L⇒lim⁡x→af(x)=Lg(x) \leq f(x) \leq h(x) \text{ and } \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L \Rightarrow \lim_{x \to a} f(x) = L

If f(x) is squeezed between g(x) and h(x) near a, and g and h have the same limit L, then f also has limit L.

Worked examples

Find lim(x→0) x² sin(1/x).
  1. We know -1 ≤ sin(1/x) ≤ 1 for all x ≠ 0
  2. Multiply by x²: -x² ≤ x² sin(1/x) ≤ x²
  3. lim(x→0) -x² = 0 and lim(x→0) x² = 0
  4. By Squeeze Theorem, lim(x→0) x² sin(1/x) = 0

Answer: 0

Find lim(x→∞) sin(x)/x.
  1. -1 ≤ sin(x) ≤ 1 for all x
  2. Divide by x (x > 0): -1/x ≤ sin(x)/x ≤ 1/x
  3. lim(x→∞) -1/x = 0 and lim(x→∞) 1/x = 0

Answer: 0

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