The limit of \sin(h)/h at h=0

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Section B.1 The limit of $\sin(h)/h$ at $h=0$

The limit $\ds \lim_{h \to 0} \frac{\sin(h)}{h}$ is $1\text{.}$ To see this first observe that the funtion $\sin(h)/h$ is even. So we just need to argue the right limit is $1\text{.}$ To that end consider the areas of the two right triangles and the sector of the unit circle in between.

Unit circles and two triangles — Figure B.1. A `diagram` for computing the limit $\ds \lim_{h\to 0} \sin(h)/h$

We have

\begin{equation*} \frac{1}{2}\sin(h)\cos(h) \le \frac{1}{2}h \le \frac{1}{2}\tan(h). \end{equation*}

Canceling the factor $1/2$ and rearranging the terms, we get

\begin{equation*} \cos(h) \le \frac{\sin{h}}{h} \le \frac{1}{\cos(h)}. \end{equation*}

Since $\cos(h) \to 1$ as $h\to 0\text{,}$ we conclude that $\ds \lim_{h \to 0} \frac{\sin(h)}{h} = 1\text{.}$

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Calculus: An intuitive approach

Section B.1 The limit of \(\sin(h)/h\) at \(h=0\)