Limit of Tan X over X at Zero
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Theorem
- $\ds \lim_{x \mathop \to 0} \frac {\tan x} x = 1$
Proof 1
By L'Hôpital's Rule:
| \(\ds \lim_{x \mathop \to 0} \frac {\tan x} x\) | \(=\) | \(\ds \lim_{x \mathop \to 0} \frac {\sec^2 x} 1\) | Derivative of Tangent Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) | Secant of Zero |
$\blacksquare$
Proof 2
| \(\ds \lim_{x \mathop \to 0} \frac {\tan x} x\) | \(=\) | \(\ds \lim_{x \mathop \to 0} \frac 1 {\cos x} \frac {\sin x} x\) | Definition of Tangent Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{x \mathop \to 0} \frac 1 {\cos x} \lim_{x \mathop \to 0} \frac {\sin x} x\) | Product Rule for Limits of Real Functions | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{x \mathop \to 0} \frac {\sin x} x\) | Cosine of Zero is One | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) | Limit of $\dfrac {\sin x} x$ at Zero |
$\blacksquare$
Proof 3
Let $f$ be the real function defined as:
- $\map f x = \sin x$
Let:
- $c = \pi$
- $h \in \openint 1 {\dfrac \pi 2}$
We have:
| \(\ds \map f {c + h}\) | \(=\) | \(\ds \map \sin {\pi + h}\) | ||||||||||||
| \(\ds \map f c\) | \(=\) | \(\ds \sin \pi\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \exists \theta \in \openint 0 1: \, \) | \(\ds \dfrac {\map \sin {\pi + h} - \sin \pi} h\) | \(=\) | \(\ds \map \cos {\pi + \theta h}\) | Mean Value Theorem | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds -\sin h - 1\) | \(=\) | \(\ds -h \cos \theta h\) | Sine of Angle plus Straight Angle, Cosine of Angle plus Straight Angle |
Hence:
| \(\ds \leadsto \ \ \) | \(\ds -h\) | \(<\) | \(\, \ds -\sin h \, \) | \(\, \ds < \, \) | \(\ds -h \cos h\) | as $0 < \theta < 1$ | ||||||||
| \(\ds \leadsto \ \ \) | \(\ds 1\) | \(<\) | \(\, \ds \dfrac {\tan h} h \, \) | \(\, \ds < \, \) | \(\ds \sec h\) |
Hence by the Squeeze Theorem for Functions:
- $\ds \lim_{h \mathop \to 0} \dfrac {\tan h} h = 1$
The result follows on renaming $h$ to $x$.
$\blacksquare$