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00000000000000000000 (to twenty places). The numerical evidence overwhelmingly suggests that the slope of the tangent line is exactly −2 and thus that it has equation y = −2x − 1. The graph of this line and y = f (x) are shown next. 037: Given: f (x) = x3 and a = 2. We computed f (a + h) − f (a − h) 2h (1) for h = 10−1 , 10−2 , . . , 10−10 . 000001, . . 00000000000000000001. The numerical evidence overwhelmingly suggests that the slope of the tangent line is 12 and thus that it has equation y = 12x − 16.

The behavior of f near −2 is best described by observing that lim x→−2+ 1 − x2 = −∞ x+2 and lim − x→−2 1 − x2 = +∞. 054: The right-hand and left-hand limits fail to exist at a = 5. If x is close to 5 but x = 5, then x − 5 is close to zero, so that (x − 5)2 is a positive number still very close to zero. Its reciprocal is therefore a very large positive number. That is, lim x→5+ 1 1 = lim− = +∞. (x − 5)2 x→5 (x − 5)2 6 (1) Unlike the previous problems of this sort, we may in this case also write lim x→5 1 = +∞.

Therefore, by the squeeze law, 1 lim x2 cos √ = 0. 028: Because −1 sin x 1 for all x, − √ 3 √ 3 x x sin 1 x √ 3 x √ √ for all x = 0. 029: lim x→0+ 3− √ x =3− x sin 1 = 0. x lim x = 3 − 0 = 3. 031: lim x→0+ lim x→1− 4 + 3x3/2 = 4 + 3 · √ = 4 − 3 · 0 = 4. lim x + x→0 x − 1 does not exist because if x < 1, then x − 1 < 0. 032: Because √ √ x → 4 , x < 4, so that and lim 4 − x = 4 − 4 = 0. − √ 4 − x is defined for all such x. 033: Because x → 2+ , x > 2, so that x2 > 4. Hence √ 2 lim+ x − 4 = 4 − 4 = 0.