Limit of Tan X over X at Zero/Proof 3

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Theorem

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


Proof

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 $x$.

$\blacksquare$


Sources

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