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

Updated March 25, 2021

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 ($x$ radians out of $2\pi$ 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 00, cosx\cos{x} tends to 11 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.