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

\(\displaystyle \sin x\)

\(=\)

\(\displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}\)

Definition of Real Sine Function

\(\displaystyle \)

\(=\)

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

\(\displaystyle \)

\(=\)

\(\displaystyle x + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}\)

\(\displaystyle \lim_{x \mathop \to 0} \frac {\sin x} x\)

\(=\)

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

\(\displaystyle \)

\(=\)

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

\(\displaystyle \)

\(=\)

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

Power Series is Differentiable on Interval of Convergence and L'Hôpital's Rule

\(\displaystyle \)

\(=\)

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

\(\displaystyle \)

\(=\)

\(\displaystyle 1 + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {0^{2 n} } {\paren {2 n}!}\)

Real Polynomial Function is Continuous

\(\displaystyle \)

\(=\)

\(\displaystyle 1\)

$\blacksquare$

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:

\(\displaystyle O\)

\(=\)

\(\displaystyle \left({0, 0}\right)\)

\(\displaystyle A\)

\(=\)

\(\displaystyle \left({1, 0}\right)\)

\(\displaystyle B\)

\(=\)

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

\(\displaystyle C\)

\(=\)

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

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:

\(\displaystyle \left \vert{\frac 1 2 \sin \theta}\right\vert\)

\(\le\)

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

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

\(\displaystyle \implies \ \ \)

\(\displaystyle \frac 1 2 \left \vert{\sin \theta}\right\vert\)

\(\le\)

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

\(\displaystyle \implies \ \ \)

\(\displaystyle 1\)

\(\le\)

\(\displaystyle \frac {\vert\theta\vert} {\vert \sin \theta\vert} \le \frac 1 {\vert \cos\theta \vert}\)

by multiplying all terms by $\dfrac 2 {\vert \sin \theta \vert}$

\(\displaystyle \implies \ \ \)

\(\displaystyle 1\)

\(\le\)

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

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): $\S 16.3 \ (4) \ \text {(i)}$