→ dogy007 :想起國中分解因式的加一項減一項了嗎 :-) 11/06 19:41
→ iamsheep :這個解法好難懂 11/06 22:00
推 ntust661 :DOG大大已經把寫法解很白了阿@@ 11/06 22:26
推 jacky7987 :sheep 其實可能只是不熟悉作法吧 慢慢習慣就好:D 11/06 22:36
※ 引述《iamsheep (??)》之銘言：
: 1.Suppose f'(1) = 1. Find the limit lim [f(x^2)-f(1)]/(x+1)
: x->-1
: ans:-2
: 2.Suppose f(1) = -1 and f'(1) = 2.
: Find the limit lim [x^2f(1)-f(x^2)]/(x-1)
: x->1
: ans:-6
: 3.given that f(2) = 1 and f'(2) = 0,Find the limit
: lim [xf(2)-2f(x)]/(x-2)
: x->2
: ans:1
: 請問怎麼解?
2. lim (x^2f(1)-f(x^2))/(x-1)
= lim f(1)(x^2-1)/(x-1) + lim(f(1)-f(x^2))/(x-1)
= ...
3. lim (xf(2)-2f(x))/(x-2)
= lim f(2)(x-2)/(x-2) + 2(f(2)-f(x))/(x-2)
= ...
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