lim x→ 0 sin x 1 − cos x In this problem attempt to evaluate: sin x x→0 1 − cos x lim using approximation. a) Substitute linear approximations for sin x and cos x into this expression. Can you tell what happens in the limit? b) Substitute quadratic approximations for sin x and cos x into this expression. Can you tell what happens in the limit? Solution sin x , the result is of the form 00 . We 1 − cos x can use approximation to study this interesting limit; the results are similar to those we get from l’Hôpital’s rule. If we replace x by 0 in the quotient a) Substitute linear approximations for sin x and cos x into this expression. Can you tell what happens in the limit? Recall that the quadratic approximations of sin x and cos x near x = 0 are: cos x ≈ sin x 1 ≈ x. Thus, lim x→0 sin x 1 − cos x x 1−1 x = lim . x→ 0 0 ≈ lim x→ 0 Replacing cos x by its linear approximation makes the denominator equal to 0. We begin to have some idea that this rational function is badly behaved, but it’s hard to draw a conclusion from this calculation. b) Substitute quadratic approximations for sin x and cos x into this expression. Can you tell what happens in the limit? The quadratic approximations of sin x and cos x near x = 0 are: 1 cos x ≈ 1 − x2 2 sin x ≈ x. 1 So: lim x→0 sin x 1 − cos x x 1 − (1 − 12 x2 ) x = lim 1 2 x→0 x 2 2 = lim . x→0 x ≈ lim x→0 sin x ap 1 − cos x proaches 0 more rapidly than the numerator, and so the value of the rational function “blows up” near x = 0. We can now see that as x approaches 0 the denominator of 2