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2.5 THE PINCHING THEOREM; TRIGONOMETRIC LIMITS 103 2.5 THE PINCHING THEOREM; TRIGONOMETRIC LIMITS Figure 2.5.1 shows the graphs of three functions f , g , h . Suppose that, as suggested by the figure, for x close to c , f is trapped between g and h . (The values of these functions at c itself are irrelevant.) If, as x tends to c , both g ( x ) and h ( x ) tend to the same limit L , then f ( x ) also tends to L . This idea is made precise in what we call the pinching theorem . y x c f g g L h h f Figure 2.5.1 THEOREM 2.5.1 THE PINCHING THEOREM Let p > 0. Suppose that, for all x such that 0 < | x c | < p , h ( x ) f ( x ) g ( x ). If lim x c h ( x ) = L and lim x c g ( x ) = L , then lim x c f ( x ) = L . PROOF Let > 0. Let p > 0 be such that if 0 < | x c | < p , then h ( x ) f ( x ) g ( x ). Choose δ 1 > 0 such that if 0 < | x c | < δ 1 , then L < h ( x ) < L + . Choose δ 2 > 0 such that if 0 < | x c | < δ 2 , then L < g ( x ) < L + . Let δ = min { p , δ 1 , δ 2 } . For x satisfying 0 < | x c | < δ , we have L < h ( x ) f ( x ) g ( x ) < L + ,
104 CHAPTER 2 LIMITS AND CONTINUITY and thus | f ( x ) L | < . Remark With straightforward modifications, the pinching theorem also holds for one-sided limits. We do not spell out the details here because we will be working with two-sided limits throughout this section. We come now to some trigonometric limits. All calculations are based on radian measure. As our first application of the pinching theorem, we prove that (2.5.2) lim x 0 sin x = 0. PROOF To follow the argument, see Figure 2.5.2. For small x = 0 0 < | sin x | = length of BP < length of AP < length of AP = | x | . Thus, for such x 0 < | sin x | < | x | . O P B A x > 0 O B A P x < 0 1 1 sin x sin x x x Figure 2.5.2 Since lim x 0 0 = 0 and lim x 0 | x | = 0, we know from the pinching theorem that lim x 0 | sin x | = 0 and therefore lim x 0 sin x = 0. From this it follows readily that (2.5.3) lim x 0 cos x = 1. † Recall that in a circle of radius 1, a central angle of x radians subtends an arc of length | x | .
2.5 THE PINCHING THEOREM; TRIGONOMETRIC LIMITS 105 PROOF In general, cos 2 x + sin 2 x = 1. For x close to 0, the cosine is positive and we have cos x = 1 sin 2 x . As x tends to 0, sin x tends to 0, sin 2 x tends to 0, and therefore cos x tends to 1. Next we show that the sine and cosine functions are everywhere continuous; which is to say, for all real numbers c , (2.5.4) lim x c sin x = sin c and lim x c cos x = cos c .