Limit of Sine of X over X
Theorem
- $\displaystyle \lim_{x \mathop \to 0} \frac {\sin x} x = 1$
Corollary
- $\displaystyle \lim_{x \mathop \to 0} \frac x {\sin x} = 1$
Proof 1
| \(\displaystyle \sin x\) | \(=\) | \(\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2n+1} }{\left({2n+1}\right)!}\) | $\quad$ Definition of the Sine Function | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \left({-1}\right)^0 \frac{x^{2 \cdot 0 + 1} } { \left({2 \cdot 0 + 1}\right)!} + \sum_{n \mathop = 1}^\infty \left({-1}\right)^n \frac {x^{2n+1} } {\left({2n+1}\right)!}\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle x + \sum_{n \mathop = 1}^\infty \left({-1}\right)^n \frac {x^{2n+1} }{\left({2n+1}\right)!}\) | $\quad$ | $\quad$ |
| \(\displaystyle \lim_{x \mathop \to 0}\frac{\sin x} x\) | \(=\) | \(\displaystyle \lim_{x \mathop \to 0} \frac{x + \sum_{n \mathop = 1}^\infty \left({-1}\right)^n \frac {x^{2n+1} }{\left({2n+1}\right)!} } x\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \lim_{x \mathop \to 0} \frac x x + \lim_{x \mathop \to 0} \frac{\sum_{n \mathop = 1}^\infty \left({-1}\right)^n \frac {x^{2n + 1} }{\left({2n+1}\right)!} } x\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle 1 + \lim_{x \mathop \to 0} \frac{\sum_{n \mathop = 1}^\infty \left({-1}\right)^n \frac {x^{2n} } {\left({2n}\right)!} } 1\) | $\quad$ Power Series is Differentiable on Interval of Convergence and L'Hôpital's Rule | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle 1 + \lim_{x \mathop \to 0} \sum_{n \mathop = 1}^\infty \left({-1}\right)^n \frac {x^{2n} }{\left({2n}\right)!}\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle 1 + \sum_{n \mathop = 1}^\infty \left({-1}\right)^n \frac {0^{2n} }{\left({2n}\right)!}\) | $\quad$ Polynomial is Continuous | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle 1\) | $\quad$ | $\quad$ |
$\blacksquare$
Proof 2
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:
- $\displaystyle \lim_{x \mathop \to 0} \frac {\sin x} x = \lim_{x \mathop \to 0} \frac {D_x \left({\sin x}\right)} {D_x \left({x}\right)} = \lim_{x \mathop \to 0} \frac {\cos x} 1 = \frac 1 1 = 1$
$\blacksquare$
Geometric Proof
Let $\theta$ be an angle in the unit circle, measured in radians.
Define the following points:
| \(\displaystyle O\) | \(=\) | \(\displaystyle \left({0, 0}\right)\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle A\) | \(=\) | \(\displaystyle \left({1, 0}\right)\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle B\) | \(=\) | \(\displaystyle \left({\cos \theta, \sin \theta}\right)\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle C\) | \(=\) | \(\displaystyle \left({1, \tan \theta}\right)\) | $\quad$ | $\quad$ |
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.
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The inequality $(A)$, then, will hold in quadrant $\text{IV}$ if the absolute value of all terms is taken, and so:
| \(\displaystyle \left \vert{\frac 1 2 \sin \theta}\right\vert\) | \(\le\) | \(\displaystyle \left\vert{\frac 1 2 \theta}\right\vert \le \left\vert{\frac 1 2 \tan \theta}\right\vert\) | $\quad$ for all $\theta \in \left({-\dfrac \pi 2 \,.\,.\, 0}\right) \cup \left({0 \,.\,.\, \dfrac \pi 2}\right)$ | $\quad$ | |||||||||
| \(\displaystyle \implies \ \ \) | \(\displaystyle \frac 1 2 \left \vert{\sin \theta}\right\vert\) | \(\le\) | \(\displaystyle \frac 1 2 \left\vert{\theta}\right\vert \le \frac 1 2 \left\vert{\tan \theta}\right\vert\) | $\quad$ | $\quad$ | ||||||||
| \(\displaystyle \implies \ \ \) | \(\displaystyle 1\) | \(\le\) | \(\displaystyle \frac {\vert\theta\vert} {\vert \sin \theta\vert} \le \frac 1 {\vert \cos\theta \vert}\) | $\quad$ by multiplying all terms by $\dfrac 2 {\vert \sin \theta \vert}$ | $\quad$ | ||||||||
| \(\displaystyle \implies \ \ \) | \(\displaystyle 1\) | \(\le\) | \(\displaystyle \left \vert{ \frac {\theta}{\sin \theta} }\right \vert \le \left \vert{ \frac 1 {\cos \theta} }\right \vert\) | $\quad$ | $\quad$ |
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
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 16.3 \ (4) \ \text {(i)}$
