CalcRef

Limits & Continuity · Special Limits

Limit of sin(x)/x

lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1

One of the most important special limits in calculus. Often proved using the Squeeze Theorem with geometric arguments.

Worked examples

Find lim(x→0) sin(3x)/x.
  1. Multiply numerator and denominator by 3: sin(3x)/x = 3 · sin(3x)/(3x)
  2. Let u = 3x. As x→0, u→0. So lim = 3 · lim(u→0) sin(u)/u = 3 · 1

Answer: 3

Find lim(x→0) sin(5x)/sin(2x).
  1. Rewrite as [sin(5x)/(5x)] · (5x) / {[sin(2x)/(2x)] · (2x)}
  2. = [sin(5x)/(5x)] / [sin(2x)/(2x)] · (5/2)
  3. Both bracketed limits → 1 as x→0

Answer: 5/2

Related formulas

Limit of (cos x - 1)/x

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