# Limit of Sine of X over X at Zero

## Theorem

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

### Corollary

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

## Proof 1

 $\ds \sin x$ $=$ $\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$ Definition of Real Sine Function $\ds$ $=$ $\ds \left({-1}\right)^0 \frac{x^{2 \cdot 0 + 1} } { \left({2 \cdot 0 + 1}\right)!} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$ $\ds$ $=$ $\ds x + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$

 $\ds \lim_{x \mathop \to 0} \frac {\sin x} x$ $=$ $\ds \lim_{x \mathop \to 0} \frac {x + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} } x$ $\ds$ $=$ $\ds \lim_{x \mathop \to 0} \frac x x + \lim_{x \mathop \to 0} \frac{\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} } x$ $\ds$ $=$ $\ds 1 + \lim_{x \mathop \to 0} \frac {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} } 1$ Power Series is Differentiable on Interval of Convergence and L'Hôpital's Rule $\ds$ $=$ $\ds 1 + \lim_{x \mathop \to 0} \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$ $\ds$ $=$ $\ds 1 + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {0^{2 n} } {\paren {2 n}!}$ Real Polynomial Function is Continuous $\ds$ $=$ $\ds 1$

$\blacksquare$

## Proof 2

From Sine of Zero is Zero:

$\sin 0 = 0$
$\map {D_x} {\sin x} = \cos x$

Then by Cosine of Zero is One:

$\cos 0 = 1$
$\map {D_x} x = 1$

Thus L'Hôpital's Rule applies and so:

$\ds \lim_{x \mathop \to 0} \frac {\sin x} x = \lim_{x \mathop \to 0} \frac {\map {D_x} {\sin x} } {\map {D_x} x} = \lim_{x \mathop \to 0} \frac {\cos x} 1 = \frac 1 1 = 1$

$\blacksquare$

## Geometric Proof

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

Define the following points:

 $\ds O$ $=$ $\ds \tuple {0, 0}$ $\ds A$ $=$ $\ds \tuple {1, 0}$ $\ds B$ $=$ $\ds \tuple {\cos \theta, \sin \theta}$ $\ds C$ $=$ $\ds \tuple {1, \tan \theta}$

and consider all $\theta$ in the open interval $\openint 0 {\dfrac \pi 2}$. 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:

$\Area \triangle OAB = \dfrac 1 2 \sin \theta$

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 $\theta \in \openint 0 {\dfrac \pi 2}$.

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:

$\Area \triangle OAC = \dfrac 1 2 \tan \theta$

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

We now have the following compound inequality:

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

for all $\theta \in \openint 0 {\dfrac \pi 2}$.

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

 $\ds \size {\frac 1 2 \sin \theta}$ $\le$ $\ds \size {\frac 1 2 \theta} \le \size {\frac 1 2 \tan \theta}$ for all $\theta \in \openint {-\dfrac \pi 2} 0 \cup \openint 0 {\dfrac \pi 2}$ $\ds \leadsto \ \$ $\ds \frac 1 2 \size {\sin \theta}$ $\le$ $\ds \frac 1 2 \size \theta \le \frac 1 2 \size {\tan \theta}$ $\ds \leadsto \ \$ $\ds 1$ $\le$ $\ds \frac {\size \theta} {\size {\sin \theta} } \le \frac 1 {\size {\cos \theta} }$ multiplying all terms by $\dfrac 2 {\size {\sin \theta} }$ $\ds \leadsto \ \$ $\ds 1$ $\le$ $\ds \size {\frac \theta {\sin \theta} } \le \size {\frac 1 {\cos \theta} }$

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

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

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 $\theta \in \openint {-\dfrac \pi 2} 0 \cup \openint 0 {\dfrac \pi 2}$.

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

for all $\theta \in \openint {-\dfrac \pi 2} 0 \cup \openint 0 {\dfrac \pi 2}$.

Taking to the limit:

$\ds \lim_{\theta \mathop \to 0} 1 = 1$
$\ds \lim_{\theta \mathop \to 0} \cos \theta = 1$

So by the Squeeze Theorem:

$\ds \lim_{\theta \mathop \to 0} \frac {\sin \theta} \theta = 1$

$\blacksquare$