Re: [微分] 極限

[keith291 (keith)]

※ 引述《debdeb ()》之銘言：
:    Find c so that
:              x+2c
:     lim     ------  = 8
:    x→-∞    x-c
:   麻煩指點一下謝謝！


let c=c(x)

1.if (x+2c)/(x-c) is 0/0 or ∞/∞ type

              x+2c           1+2c'
     lim     ------ =  lim  -------
    x→-∞    x-c    x→-∞   1-c'



if (1+2c')/(1-c') is 0/0 or ∞/∞ type

we have 2c"/-c"=-2≠8  (→←)



let (1+2c')/(1-c')=8

1+2c'=8-8c'

c(x)=(7/10)x + C



2.if  (x+2c)/(x-c)=8

c(x)=(7/10)x which is included in answer c(x)=(7/10)x + C


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推 TownMayor:grt.
→ yhliu:若 c 是常數,無解; 若 c 不是常數,無窮多解.
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推 debdeb:謝謝！！