Take it to the limit

A famous limit

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

Proof

We will show that $\forall \epsilon \gt 0$ and $\forall x \neq 0$, $\exists \delta \gt 0$ such that:

$|x| \lt \delta \Longrightarrow |\frac{\sin x}{x} - 1| \lt \epsilon$.

Assumptions

Let $\epsilon \gt 0$ be given.

Let $0 \lt |x| \lt \delta$, where $\delta = \min\{\frac{\pi}{2}, \sqrt{2 \epsilon}\}$.

Then:

Since $\delta = \min\{\frac{\pi}{2},\sqrt{2 \epsilon}\}$,
$|x| \lt \delta$ implies that $|x| \lt \frac{\pi}{2}$

Note from the definition of $\sin$, without loss of generality for angles $x \in (-\frac{\pi}{2},0) \cup (0, \frac{\pi}{2})$, $|x|$ is the distance of an arc on a circle, while $\sin |x|$ is the altitude, from the x-axis, of a triangle whose hypoteneuse the radius not along the x-asix. Since an altitude is less than a hypoteneuse and a chord is less than the arc that subtends it, we have:

$\sin |x| \lt |x|$

We will also need the fact that on this interval $\tan |x| \gt |x|$. Recall that on the unit circle, $\tan |x|$ is the line segment tangent to the circle from the circle's intersection with $\sin |x|$ to the x-axis. Moreover this tangent line is perpendicular to the circle's radius, 1. As such it forms the base of a right triangle with height 1, while the sector subtended by angle $x$ lies inside that triangle. Now we proceed:

• Recalling the formula for the area of a sector:

  $\frac{1}{2}(r^2) |x|$, where $x$ is the angle.

  Then, we have that the sector carved out by the angle x is smaller than the triangle formed by using the radius and tangent x as legs:

  $\frac{1}{2}(1^2) |x| \lt \frac{1}{2}(1) \tan |x|$

  or

  $|x| \lt \tan |x|$.

• From there we derive:

  $\cos |x| \lt \frac{\sin |x|}{|x|}$,

• recalling from above that $\sin |x| \lt |x|$ and noting that $\sin$ is an odd function, we have:

  $\cos |x| \lt \frac{\sin x}{x} \lt 1$

• $\cos |x| - 1 \lt \frac{\sin x}{x} - 1 \lt 0$

• And since $1 - \cos |x| \gt 0$, we have:

  $|\frac{\sin x}{x} - 1| \lt 1 - \cos |x|$

• But $1 - \cos |x| = 2 \sin^2 \frac{x}{2}$, and $\sin|x| \lt |x| \lt \delta \rightarrow 2 \sin^2 \frac{x}{2} \lt \frac{\delta^2}{2}$ so:

  $|\frac{\sin x}{x} - 1| \lt 2 \sin^2 \frac{x}{2} \lt \frac{\delta^2}{2}$,

Conclusion

By picking $\delta = \min\{\frac{\pi}{2},\sqrt{2 \epsilon}\}$,

$|x| \lt \delta \rightarrow |\frac{\sin x}{x} - 1| \lt \frac{\delta^2}{2} \leq \epsilon$.

As desired. $\blacksquare$