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:

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

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

$\blacksquare$

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:

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

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

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

$b = 1$

$h = \sin \theta$

and so:

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

From Area of a Sector, the sector formed by arc $BC$ subtending $A$ is $\dfrac \theta 2$.

Clearly this sector cannot be smaller in area than $\triangle ABC$, 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 a Triangle in Terms of Side and Altitude, we have that $\triangle ABD$ has an area of $\dfrac 1 2 b h$ where:

$b = 1$

$h = \tan \theta$

and so:

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

$\triangle ABD$ 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 $\theta \in \left({0 . . \dfrac \pi 2}\right)$.

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 IV if the absolute value of all terms is taken, and so:

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

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:

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

$\blacksquare$

Proof of Corollary

We have the inequality

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

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

Taking the limit of the leftmost term and the rightmost term:

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

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

So by the Squeeze Theorem:

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

$\blacksquare$

Sources

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