CLP-1 Differential Calculus

Subsection 2.8.2 Step 1: \(\mathbf{\diff{}{x} \{ \sin x \} \big|_{x=0}}\)

By definition, the derivative of \(\sin x\) evaluated at \(x=0\) is

We will prove this limit by use of the squeeze theorem (Theorem 1.4.18). To get there we will first need to do some geometry. But first we will build some intuition.

The figure below contains part of a circle of radius 1. Recall that an arc of length \(h\) on such a circle subtends an angle of \(h\) radians at the centre of the circle. So the darkened arc in the figure has length \(h\) and the darkened vertical line in the figure has length \(\sin h\text{.}\) We must determine what happens to the ratio of the lengths of the darkened vertical line and darkened arc as \(h\) tends to zero.

Here is a magnified version of the part of the above figure that contains the darkened arc and vertical line.

This particular figure has been drawn with \(h=.4\) radians. Here are three more such blow ups. In each successive figure, the value of \(h\) is smaller. To make the figures clearer, the degree of magnification was increased each time \(h\) was decreased.

As we make \(h\) smaller and smaller and look at the figure with ever increasing magnification, the arc of length \(h\) and vertical line of length \(\sin h\) look more and more alike. We would guess from this that

The following tables of values

suggests the same guess. Here is an argument that shows that the guess really is correct.