2.3 The limit laws  (Page 5/15)

Because $\underset{\theta \to 0^{+}}{\text{lim}}0=0$ and $\underset{x\to 0^{+}}{\text{lim}}\theta =0,$ 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 [link] . Notice that this figure adds one additional triangle to [link] . 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 [link] 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.