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Chapter 2 | Limits
Figure 2.29 The sine function is shown as a line on the unit circle.
Because lim θ →0 +
0=0 and lim θ →0 +
θ =0, by using the squeeze theorem we conclude that
lim θ →0 +
sin θ =0.
To see that and hence, 0< sin(− θ ) <− θ . Consequently, 0< −sin θ <− θ . It follows that 0> sin θ > θ . An application of the squeeze theorem produces the desired limit. Thus, since lim θ →0 + sin θ =0 and lim θ →0 − sin θ =0, (2.16) lim θ →0 sin θ =0. Next, using the identity cos θ = 1−sin 2 θ for − π 2 < θ < π 2 , we see that (2.17) lim θ →0 cos θ = lim θ →0 1−sin 2 θ =1. We now take a look at a limit that plays an important role in later chapters—namely, lim θ →0 sin θ θ . 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 tan θ . Thus, we see that for 0< θ < π 2 , sin θ < θ < tan θ . lim θ →0 − sin θ =0 as well, observe that for − π 2 < θ <0, 0<− θ < π 2
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