Proof that limit as x → 0 of sin(x)/x = 1


limx0sinxx=1\lim_{x \to 0} \frac{\sin x}{x} = 1

This is a unit circle, so there two radii are labeled with length 11. The vertical line labeled sinx\sin{x} because sine of the angle xx is that line over the hypotenuse (11). The second vertical line is labeled tanx\tan{x} because tangent of the angle xx is that line over the base of the triangle (11).

Compare the areas of three regions on the diagram above:

  • Blue: 12(base)(height)=12(1)(sinx)=12sinx\frac{1}{2}(\text{base})(\text{height}) = \frac{1}{2}(1)(\sin{x}) = \frac{1}{2}\sin{x}
  • Blue + yellow: x2π(π)=xπ2π=x2\frac{x}{2\pi}(\pi) = \frac{x\pi}{2\pi} = \frac{x}{2}
    • The area of that sector of the circle is that part of the circle (xx radians out of 2π2\pi total radians around the circle) times the area of the circle (πr2\pi r^2, or π12\pi 1^2, so π\pi).
  • Blue + yellow + red: 12(base)(height)=12(1)(tanx)=12tanx\frac{1}{2}(\text{base})(\text{height}) = \frac{1}{2}(1)(\tan{x}) = \frac{1}{2}\tan{x}

We can see visually that these areas can be arranged in order of size like this:

  • 12sinx<x2<12tanx\frac{1}{2} \sin{x} < \frac{x}{2} < \frac{1}{2} \tan{x}
  • sinx<x<tanx\sin{x} < x < \tan{x}
  • sinxsinx<xsinx<tanxsinx\frac{\sin{x}}{\sin{x}} < \frac{x}{\sin{x}} < \frac{\tan{x}}{\sin{x}}
    • Divide each term by sinx\sin{x}
  • 1<xsinx<1cosx1 < \frac{x}{\sin{x}} < \frac{1}{\cos{x}}
    • Simplify the last step
  • 1>sinxx>cosx1 > \frac{\sin{x}}{x} > \cos{x}
    • Take the reciprocal of each term, and therefore flip inequality signs

By the Squeeze Theorem, we can see that sinxx\frac{\sin{x}}{x} is "squeezed" between 11 and cosx\cos{x}. As xx tends to 0, cosx\cos{x} tends to 1 as well.

Therefore, sinxx\frac{\sin{x}}{x} is squeezed between 1 and 1 as x → 0, so the limit of sinxx\frac{\sin{x}}{x} is 1.

This page is referenced in: Grade 12, Limits and Continuity