Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 51

Below is the figure of a circular arc of length $s$ and a chord of length $d$ both subtended by a central angle $\theta$. Find $\displaystyle \lim\limits_{\theta \to 0^+}\frac{s}{d}$




We will use the formula for arc to make $s$ in terms of $r \text{ and } \theta$, so...
$s = r \theta \qquad \Longleftarrow \text{ Equation1}$

Also, we can divide the triangle like this





$
\begin{equation}
\begin{aligned}
\sin \left(\frac{\theta}{2}\right) &= \frac{\frac{d}{2}}{r}\\
\\
\sin \left(\frac{\theta}{2}\right) &= \frac{d}{2r}\\
\\
d &= 2r \sin \left( \frac{\theta}2{}\right) && \Longleftarrow \text{ Equation 2}
\end{aligned}
\end{equation}
$


Plugging in Equations 1 and 2 to the limit we get,


$
\begin{equation}
\begin{aligned}
\lim\limits_{\theta \to 0^+} \frac{s}{d} &= \lim\limits_{\theta \to 0^+} \frac{\cancel{r}\theta}{2\cancel{r}\sin\left(\frac{\theta}{2}\right)}\\
\\
\lim\limits_{\theta \to 0^+} \frac{s}{d} &= \lim\limits_{\theta \to 0^+} \frac{\theta}{2\sin\left(\frac{\theta}{2}\right)}\\
\\
\lim\limits_{\theta \to 0^+} \frac{s}{d} &= \lim\limits_{\theta \to 0^+} \frac{\left(\frac{1}{2}\right) \theta}{\cancel{\left(\frac{1}{2}\right)}\cancel{2}\sin\left(\frac{\theta}{2}\right)} \\
\\
&= \lim\limits_{\theta \to 0^+} \frac{\frac{\theta}{2}}{\sin \left( \frac{\theta}{2}\right)}
\end{aligned}
\end{equation}
$

Recall that $\displaystyle \lim\limits_{\theta \to 0^+} \frac{\sin \theta}{\theta}= 1$
Therefore, $\displaystyle \lim\limits_{\theta \to 0^+} \frac{s}{d} = 1$