Re: [微積] 請教一題漸進線題目
※ 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~
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03/11 15:28, , 1F
03/11 15:28, 1F
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