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Chapter 4 47

# Chapter 4 47 - SECTION 4.4 lNDETERMINATE FORMS AND...

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Unformatted text preview: SECTION 4.4 lNDETERMINATE FORMS AND L'HOSPITAL'S RULE 309 75. Since lim[f(w + h) — f(a: — h)] 2 f(;v) — ﬂag) 2 0 (f is differentiable and hence continuous) and Ilmh 2h 2 0., h—vO _. we use l‘l-lospital's Rule: . fl\$+h)—f(/_h)ﬂ. f’(m+h)(1)-f'(m-h)(21) f’(\$)+f’(\$)_2f’(\$)_l ,1.me m 2 —ﬁ 2 —f<z> f(\$+h)-f(\$—h) 2h between (x — h, f(:c — h)) and (a: + h, f(m + h)). As is the slope of the secant line h —» 0, this line gets closer to the tangent line and its slope approaches f’ (m). 76. Since gig] [f(:c + h) — 2f(.1:) + ﬂat 2 h)] 2 ﬂan) ~ 2f(:c) + f(\$) 2 0 (f is differentiable and hence continuous) and lint) h2 2 0. we can apply l’Hospital’s Rule: _ h—2 —h . ’+h—’—h ,, ,{gf(m+) 12(2me(00 )ggﬂﬂw )2hf(at )_f(\$) At the last step. we have applied the result of Exercise 75 to f /(m) 77. (a) We show that lim ﬂat) 2 0 for every integer n 2 0. Let y 2 ~1—2 Then \$20 a)” J} —1/91:2 n n—l l 1imf(m)=1‘e,n:limy—§hmny i-énmlﬂ): z—>O 1‘2" :I:—>O (\$2) y—>oo 6y yaoo 6y y—>oo ell 11m L”) = lim x" Hm) : lim x" lim fl” = 0. Thus. f’(0) : lim L” T “0) : li L95) : 0. 2—0 1:” z—>0 51:27" z—>O m—>0 \$2" 32—0 1: — 0 z—>O {I} (b) Using the Chain Rule and the Quotient Rule we see that f (n) (9c) exists for 2: 7S 0. In fact. we prove by induction that for each n 2 0. there is a polynomial p" and a non—negative integer kn with f(")(ac) 2 pn(:c)f(x)/:ck" for m 2 0. This is true for n 2 0; suppose it is true for the nth derivative. Then f'(m) 2 f(m)(2/ac3) so f‘"+”<x> : [W [p:.(m>f(m> + Woman] — knxk"’1pn(:c)f(w)lm‘2k" = lmka) Mme/\$3) — lam—1mm] fan-2k” = lw’“"+3p£.(x) + 2mm) ~ knprnm] f(r)x“2'“"+3) which has the desired form. Now we show by induction that f(") (O) 2 0 for all 71. By part (a). f’(0) 2 0. Suppose that fl")(0) 2 0. Then (11) _ (n) (n) kn f<n+n(0):hmf (x) f (0):hmf (w):hm memo/a: :hm WW) 2:20 zy—O \$20 :1: \$2.0 a» z_,0 xkn+1 . . x ﬁancee) .1135 9325.3. = m0) . 0 z 0 ...
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