. Differential and integral calculus, an introductory course for colleges and engineering schools. §11 LIMITS 21 Example 3. To determine lim • 0=0 d Solution. If in this function we write 0 for 9, the function takes the meaningless form -. Hence we cannot here determine the limit by simply substituting the limit of the argument in the place of the argument, ascould have been done in the precedingexamples. In the unit circle consideran arc AC = 9, which for convenience we suppose to lie between 0 and |- AD, BC, and OB are respectively tan 9,sin 9, and cos 9. From the figure it isobvious that ar
. Differential and integral calculus, an introductory course for colleges and engineering schools. §11 LIMITS 21 Example 3. To determine lim • 0=0 d Solution. If in this function we write 0 for 9, the function takes the meaningless form -. Hence we cannot here determine the limit by simply substituting the limit of the argument in the place of the argument, ascould have been done in the precedingexamples. In the unit circle consideran arc AC = 9, which for convenience we suppose to lie between 0 and |- AD, BC, and OB are respectively tan 9,sin 9, and cos 9. From the figure it isobvious that area OAD > area OAC > area geometry Hence. area OAD = \ tan 0; area OAC- = i 0; area OBC = \ sin 9 cos 9. tan 9 > 9 > sin 9 cos 9, > -— > COS 9, cos 9 sin 9 n ^ smflcos 9 < < cos 91 Now let 6 = 0 (C = A): then cos 9 = 1, and = 1. cos 9 Hence, since sin 9 lies always between cos 9 and , the limit of the function is also 1. cos 9 That is I» lim —3— = 1. 9=0 e Example 4. To determine lim 0=0 1 — cos 9 Solution. There is no loss of generality in taking 9 at such a stage of3 variation that either 0 <cos 9 is positive. We have its variation that either 0 < 9 0>-^,so that in either case 1— cos 1— cos2fl 1 + COS 9 sm2fl1 + cos 9 < sin2 9. 22 DIFFERENTIAL CALCULUS §12 Further, sin 6 < 6 and sin2 6 < 6 sin 6. Hence, because 1 — cos d is positive,we have i cos 0 0 < 1 — cos e < 6 sin 0, and 0 < < sin 6. e But lim sin 6 = 0; therefore, 0=0 -r-r «• 1 —cos© -IT. lim s = 0. 12. The Difference between the Limit of /(a?) when x = a,and the Value of /(«) when
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