# Limit of Sine of X over X

## Contents

## Theorem

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

### Corollary

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

## Proof

### Direct Proof from Definition of Sine

This proof works directly from the definition of the sine function:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \sin x\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n+1} }{\left({2n+1}\right)!}\) | \(\displaystyle \) | \(\displaystyle \) | By the definition of the sine function | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left({-1}\right)^0 \frac{x^{2 \cdot 0 + 1} } { \left({2 \cdot 0 + 1}\right)!} + \sum_{n=1}^\infty \left({-1}\right)^n \frac {x^{2n+1} } {\left({2n+1}\right)!}\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle x + \sum_{n=1}^\infty \left({-1}\right)^n \frac {x^{2n+1} }{\left({2n+1}\right)!}\) | \(\displaystyle \) | \(\displaystyle \) |

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \lim_{x \to 0}\frac{\sin x} x\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \lim_{x \to 0} \frac{x + \sum_{n=1}^\infty \left({-1}\right)^n \frac {x^{2n+1} }{\left({2n+1}\right)!} } x\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \lim_{x \to 0} \frac x x + \lim_{x \to 0} \frac{\sum_{n=1}^\infty \left({-1}\right)^n \frac {x^{2n + 1} }{\left({2n+1}\right)!} } x\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle 1 + \lim_{x \to 0} \frac{\sum_{n=1}^\infty \left({-1}\right)^n \frac {x^{2n} } {\left({2n}\right)!} } 1\) | \(\displaystyle \) | \(\displaystyle \) | by Power Series Differentiable on Interval of Convergence and L'Hôpital's Rule | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle 1 + \lim_{x \to 0} \sum_{n=1}^\infty \left({-1}\right)^n \frac {x^{2n} }{\left({2n}\right)!}\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle 1 + \sum_{n=1}^\infty \left({-1}\right)^n \frac {0^{2n} }{\left({2n}\right)!}\) | \(\displaystyle \) | \(\displaystyle \) | by Polynomial is Continuous | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle 1\) | \(\displaystyle \) | \(\displaystyle \) |

$\blacksquare$

## Alternative Proof

This proof assumes the truth of the Derivative of Sine Function:

We have that:

- From Sine of Zero is Zero: $\sin 0 = 0$;
- From Derivative of Sine Function: $D_x \left({\sin x}\right) = \cos x$. Then by Cosine of Zero is One, $\cos 0 = 1$;
- From Derivative of Identity Function: $D_x \left({x}\right) = 1$.

Thus L'Hôpital's Rule applies and so $\displaystyle \lim_{x \to 0} \frac {\sin x} x = \lim_{x \to 0} \frac {\cos x} 1 = \frac 1 1 = 1$.

$\blacksquare$

### Geometric Proof

Let $\theta$ be an angle in the unit circle, measured in radians.

Define the following points:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle O\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left({0, 0}\right)\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle A\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left({1, 0}\right)\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle B\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left({\cos \theta, \sin \theta}\right)\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle C\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left({1, \tan \theta}\right)\) | \(\displaystyle \) | \(\displaystyle \) |

and consider all $\theta$ in the open interval $\left({0 \,.\,.\, \dfrac \pi 2}\right)$.

From Area of Triangle in Terms of Side and Altitude, we have that $\triangle OAB$ has an area of $\dfrac 1 2 b h$ where:

- $b = 1$
- $h = \sin \theta$

and so:

- $\operatorname {area}\triangle OAB = \dfrac 1 2 \sin \theta$

From Area of Sector, the sector formed by arc $AB$ subtending $O$ is $\dfrac \theta 2$.

Clearly this sector cannot be smaller in area than $\triangle OAB$, and so we have the inequality:

- $\dfrac {\sin \theta} 2 \le \dfrac \theta 2$

for all $\theta \in \left({0 \,.\,.\, \dfrac \pi 2}\right)$.

Next, from Area of Triangle in Terms of Side and Altitude, we have that $\triangle OAC$ has an area of $\dfrac 1 2 b h$ where:

- $b = 1$
- $h = \tan \theta$

and so:

- $\operatorname {area}\triangle OAC = \dfrac 1 2 \tan \theta$

$\triangle OAC$ is clearly not smaller than the sector.

We now have the following compound inequality:

- $(A) \quad \dfrac 1 2 \sin \theta \le \dfrac 1 2 \theta \le \dfrac 1 2 \tan \theta$

for all $\theta \in \left({0 \,.\,.\, \dfrac \pi 2}\right)$.

Clearly, if any of the plane regions were to be reflected about the $x$-axis, the magnitudes of the signed areas would be the same.

The inequality $(A)$, then, will hold in quadrant $\text{IV}$ if the absolute value of all terms is taken, and so:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left \vert{\frac 1 2 \sin \theta}\right\vert\) | \(\le\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left\vert{\frac 1 2 \theta}\right\vert \le \left\vert{\frac 1 2 \tan \theta}\right\vert\) | \(\displaystyle \) | \(\displaystyle \) | for all $\theta \in \left({-\dfrac \pi 2 \,.\,.\, 0}\right) \cup \left({0 \,.\,.\, \dfrac \pi 2}\right)$ | ||

\(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle \frac 1 2 \left \vert{\sin \theta}\right\vert\) | \(\le\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \frac 1 2 \left\vert{\theta}\right\vert \le \frac 1 2 \left\vert{\tan \theta}\right\vert\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle 1\) | \(\le\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \frac {\vert\theta\vert} {\vert \sin \theta\vert} \le \frac 1 {\vert \cos\theta \vert}\) | \(\displaystyle \) | \(\displaystyle \) | by multiplying all terms by $\dfrac 2 {\vert \sin \theta \vert}$ | ||

\(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle 1\) | \(\le\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left \vert{ \frac {\theta}{\sin \theta} }\right \vert \le \left \vert{ \frac 1 {\cos \theta} }\right \vert\) | \(\displaystyle \) | \(\displaystyle \) |

Now, we have that $\dfrac{\theta}{\sin\theta} \ge 0$ on $\left({-\dfrac \pi 2 \,.\,.\, 0}\right) \cup \left({0 \,.\,.\, \dfrac \pi 2}\right)$.

Also, we have that $\dfrac 1 {\cos \theta} \ge 0$ on $\left({-\dfrac \pi 2 \,.\,.\, 0}\right) \cup \left({0 \,.\,.\, \dfrac \pi 2}\right)$.

So our absolute value signs are not needed.

Hence we arrive at:

- $1 \le \dfrac \theta {\sin \theta} \le \dfrac 1 {\cos \theta}$

for all $\theta \in \left({-\dfrac \pi 2 \,.\,.\, 0}\right) \cup \left({0 \,.\,.\, \dfrac \pi 2}\right)$.

Inverting all terms and reversing the inequalities:

- $1 \ge \dfrac{\sin\theta}{\theta} \ge \cos \theta$

for all $\theta \in \left({-\dfrac \pi 2 \,.\,.\, 0}\right) \cup \left({0 \,.\,.\, \dfrac \pi 2}\right)$.

Taking to the limit:

- $\displaystyle \lim_{\theta \to 0} \ 1 = 1$

- $\displaystyle \lim_{\theta \to 0} \ \cos \theta = 1$

So by the Squeeze Theorem:

- $\displaystyle \lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$

$\blacksquare$

## Sources

- K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*(1977)... (previous)... (next): $\S 16.3 \ (4) \ \text {(i)}$