Limit of Sine of X over X

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Theorem

$\ds \lim_{x \mathop \to 0} \frac {\sin x} x = 1$


Corollary

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


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 $\openint 0 {\dfrac \pi 2}$.

Limit of Sine of X over X-Proof 3.png

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.


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