推 viwa77068194: 謝謝你的詳解 03/11 15:28
※ 引述《viwa77068194 (Yvette)》之銘言：
: The line y=mx+b is called a slant asymptote of the graph y=f(x)
: if lim x →∞ (f(x)-(mx+b))=0 or lim x →-∞ (f(x)-(mx+b))=0
: Find all (horizontal,vertical,slant) asymptote of the graph of y=ln(1+e^x)
lim [ln(1 + e^x) - (mx + b)] = 0
x→∞
=> m = 1
lim [ln(1 + e^x) - x] = b
x→∞
=> b = 0
lim [ln(1 + e^x) - (mx + b)] = 0
x →-∞
=> m = 0
lim [ln(1 + e^x)] = b
x →-∞
=> b = 0
所以有兩條漸近線
y = x
y = 0
: 它在漸進線章節
: 謝謝回答的人~
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 61.228.131.45
※ 文章網址: https://www.ptt.cc/bbs/Math/M.1426056195.A.FCF.html