Unformatted text preview: SECTION 4.4 lNDETERMINATE FORMS AND L'HOSPITAL'S RULE 301 75. There must exrst some interval containing con which f’" is positive. since f’"(c) is positive and f’ is continuous. On this interval. fll is increasing (since f’” is positive). so f” = (f’)’ changes from negative to positive at 0. So by the First Derivative Test. f/ has a local minimum at w 2 c and thus cannot Change sign there. so f has no maximum or minimum at 0. But since f” Changes from negative to positive at c. f has a point of inﬂection at c (it changes from concave down to concave up). 4.4 Indeterminate Forms and L'Hospital's Rule \ The use of l'Hospital's Rule is indicated by an H above the equal sign: 2 . 1. (3) lim M . . . 0
IS an indeterminate form of type —.
z—nz g .73) (b) lim % : 0 because the numerator approaches 0 while the denominator becomes large. . h a: . . .
(C) 11m 1% : 0 because the numerator approaches a ﬁnite number while the denominator becomes large. (d) If lim p(a:) : 00 and ﬁx) z—aa p(:v) : 1/232. and ﬂan) : 902.] If f(m) ~> 0 through negative values. then lim Mi) — —00. For exam 1e. take a = 0,1)(12) 2 1/332. and f($) 2 —:c2.] If ﬁx) ~> 0 through both positive and negative values. then the limit might not exist. [For example. take a = O. p(:c) : UN, and f(;i:) : 32.] (e) hm M IS an indeterminate form of type E.
zHa q(w) oo 2. (a) gin; [f(x)p(;c)] is an indeterminate fonn of type 0 - 00. (b) When m is near a. p(.r) is large and Mm) is near 1. so h(:c)p(m) is large. Thus. lim [h(a:)p(a:)] : 00.
(c) When :E is near a. p(a:) and q(m) are both large, so p(m)q(m) is large. Thus. Iim [p(a:)q(x)] = 00. 3. (a) When :c is near a. f(:L') is near 0 and 10(22) is large, so f(:L-) — p(:c) is large negative. Thus.
gig; use) — pm] : —oo. (b) 1113“; [p(m) — q(m)] is an indeterminate form of type 00 — oo. ...
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