第2章極限

The techniques we have developed thus far work very well for algebraic functions, but we are still unable to evaluate limits of very basic trigonometric functions. The next theorem, called the squeeze theorem, proves very useful for establishing basic trigonometric limits. This theorem allows us to calculate limits by “squeezing” a function, with a limit at a point a that is unknown, between two functions having a common known limit at a. Figure 2.27 illustrates this idea.

We now use the squeeze theorem to tackle several very important limits. Although this discussion is somewhat lengthy, these limits prove invaluable for the development of the material in both the next section and the next chapter. The first of these limits is $\underset{\theta \to 0}{\text{lim}}\text{sin}\theta .$ Consider the unit circle shown in Figure 2.29. In the figure, we see that $\text{sin}\theta$ is the y-coordinate on the unit circle and it corresponds to the line segment shown in blue. The radian measure of angle θ is the length of the arc it subtends on the unit circle. Therefore, we see that for $0<\theta <\frac{\pi }{2},0<\text{sin}\theta <\theta .$

Because $\underset{\theta \to 0^{+}}{\text{lim}}0=0$ and $\underset{\theta \to 0^{+}}{\text{lim}}\theta =0,$ by using the squeeze theorem we conclude that

To see that $\underset{\theta \to 0^{-}}{\text{lim}}\text{sin}\theta =0$ as well, observe that for $-\frac{\pi }{2}<\theta <0,0<\text{−}\theta <\frac{\pi }{2}$ and hence, $0<\text{sin}\left(-\theta \right)<\text{−}\theta .$ Consequently, $0<-\text{sin}\theta <\text{−}\theta .$ It follows that $0>\text{sin}\theta >\theta .$ An application of the squeeze theorem produces the desired limit. Thus, since $\underset{\theta \to 0^{+}}{\text{lim}}\text{sin}\theta =0$ and $\underset{\theta \to 0^{-}}{\text{lim}}\text{sin}\theta =0,$

Next, using the identity $\text{cos}\theta =\sqrt{1-{\text{sin}}^2\theta }$ for $-\frac{\pi }{2}<\theta <\frac{\pi }{2},$ we see that

We now take a look at a limit that plays an important role in later chapters—namely, $\underset{\theta \to 0}{\text{lim}}\frac{\text{sin}\theta }{\theta }.$ To evaluate this limit, we use the unit circle in Figure 2.30. Notice that this figure adds one additional triangle to Figure 2.30. We see that the length of the side opposite angle θ in this new triangle is $\text{tan}\theta .$ Thus, we see that for $0<\theta <\frac{\pi }{2},\text{sin}\theta <\theta <\text{tan}\theta .$

By dividing by $\text{sin}\theta$ in all parts of the inequality, we obtain

Equivalently, we have

Since $\underset{\theta \to 0^{+}}{\text{lim}}1=1=\underset{\theta \to 0^{+}}{\text{lim}}\text{cos}\theta ,$ we conclude that $\underset{\theta \to 0^{+}}{\text{lim}}\frac{\text{sin}\theta }{\theta }=1.$ By applying a manipulation similar to that used in demonstrating that $\underset{\theta \to 0^{-}}{\text{lim}}\text{sin}\theta =0,$ we can show that $\underset{\theta \to 0^{-}}{\text{lim}}\frac{\text{sin}\theta }{\theta }=1.$ Thus,

In Example 2.25 we use this limit to establish $\underset{\theta \to 0}{\text{lim}}\frac{1-\text{cos}\theta }{\theta }=0.$ This limit also proves useful in later chapters.