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Subsection 2.8.3 Proof that \(\mathbf{\lim\limits_{h\rightarrow 0}\tfrac{\sin h}{h}=1}\)

The circle in the figure above has radius \(1\text{.}\) Hence

\begin{align*} |OP|=|OR|&=1 & |PS|&=\sin h\\ |OS|&= \cos h & |QR|&=\tan h \end{align*}

Now we can use a few geometric facts about this figure to establish both an upper bound and a lower bound on \(\frac{\sin h}{h}\) with both the upper and lower bounds tending to \(1\) as \(h\) tends to \(0\text{.}\) So the squeeze theorem will tell us that \(\frac{\sin h}{h}\) also tends to \(1\) as \(h\) tends to \(0\text{.}\)

  • The triangle \(OPR\) has base \(1\) and height \(\sin h\text{,}\) and hence
    \begin{align*} \text{area of }\triangle OPR &= \half\times1\times\sin h=\frac{\sin h}{2}. \end{align*}
  • The triangle \(OQR\) has base \(1\) and height \(\tan h\text{,}\) and hence
    \begin{align*} \text{area of }\triangle OQR &= \half\times1\times\tan h=\frac{\tan h}{2}. \end{align*}
  • The “piece of pie” \(OPR\) cut out of the circle is the fraction \(\frac{h}{2\pi}\) of the whole circle (since the angle at the corner of the piece of pie is \(h\) radians and the angle for the whole circle is \(2\pi\) radians). Since the circle has radius \(1\) we have
    \begin{align*} \text{area of pie } OPR &= \frac{h}{2\pi} \cdot (\text{area of circle}) = \frac{h}{2\pi} \pi \cdot 1^2= \frac{h}{2} \end{align*}

Now the triangle \(OPR\) is contained inside the piece of pie \(OPR\text{.}\) and so the area of the triangle is smaller than the area of the piece of pie. Similarly, the piece of pie \(OPR\) is contained inside the triangle \(OQR\text{.}\) Thus we have

\begin{gather*} \text{area of triangle } OPR \leq \text{ area of pie } OPR \leq \text{ area of triangle } OQR \end{gather*}

Substituting in the areas we worked out gives

\begin{align*} \frac{\sin h}{2} &\leq \frac{h}{2} \leq \frac{\tan h}{2}\\ \end{align*}

which cleans up to give

\begin{align*} \sin h &\leq h \leq \frac{\sin h}{\cos h} \end{align*}

We rewrite these two inequalities so that \(\frac{\sin h}{h}\) appears in both.

  • Since \(\sin h \leq h\text{,}\) we have that \(\ds \frac{\sin h}{h} \leq 1\text{.}\)
  • Since \(\ds h \leq \frac{\sin h}{\cos h}\) we have that \(\ds \cos h \leq \frac{\sin h}{h}\text{.}\)

Thus we arrive at the “squeezable” inequality

\begin{gather*} \cos h \leq \frac{\sin h}{h} \leq 1 \end{gather*}

We know  2 Again, refresh your memory by looking up Appendix A.5. that

\begin{align*} \lim_{h \to 0} \cos h &=1. \end{align*}

Since \(\tfrac{\sin h}{h}\) is sandwiched between \(\cos h\) and 1, we can apply the squeeze theorem for limits (Theorem 1.4.18) to deduce the following lemma:

Since this argument took a bit of work, perhaps we should remind ourselves why we needed it in the first place. We were computing

\begin{align*} \diff{}{x} \{ \sin x\}\Big|_{x=0} &= \lim_{h \to 0} \frac{\sin h - \sin 0}{h}\\ &= \lim_{h \to 0} \frac{\sin h}{h} & \text{(This is why!)}\\ &= 1 \end{align*}

This concludes Step 1. We now know that \(\diff{}{x}\sin x\big|_{x=0}=1\text{.}\) The remaining steps are easier.

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