Limit of Sine of X over X

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:

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

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

$\blacksquare$

Alternative Proof

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

We have that:

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:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle O$$ $$=$$ $$\displaystyle$$ $$\displaystyle \left({0, 0}\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle A$$ $$=$$ $$\displaystyle$$ $$\displaystyle \left({1, 0}\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle B$$ $$=$$ $$\displaystyle$$ $$\displaystyle \left({\cos \theta, \sin \theta}\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle C$$ $$=$$ $$\displaystyle$$ $$\displaystyle \left({1, \tan \theta}\right)$$ $$\displaystyle$$ $$\displaystyle$$

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 triple 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:

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

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

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

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:

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

$\blacksquare$