Limit of Sine of X over X
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:
- $\sin 0 = 0$
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:
| \(\ds O\) | \(=\) | \(\ds \tuple {0, 0}\) | ||||||||||||
| \(\ds A\) | \(=\) | \(\ds \tuple {1, 0}\) | ||||||||||||
| \(\ds B\) | \(=\) | \(\ds \tuple {\cos \theta, \sin \theta}\) | ||||||||||||
| \(\ds C\) | \(=\) | \(\ds \tuple {1, \tan \theta}\) |
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.
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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 $\theta \in \openint {-\dfrac \pi 2} 0 \cup \openint 0 {\dfrac \pi 2}$.
Inverting all terms and reversing the inequalities:
- $1 \ge \dfrac {\sin\theta} \theta \ge \cos \theta$
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:
- $\ds \lim_{\theta \mathop \to 0} \frac {\sin \theta} \theta = 1$
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 16.3 \ (4) \ \text {(i)}$