asymptote?

2016-04-22 2:35 pm
find the asymptote of f(x)=x^2[e^(0.5x+1)]

回答 (1)

2016-04-23 3:37 am
✔ 最佳答案
令 u = - x
當 x → - ∞ , 即 u → ∞ 時 :

f
= ( - u )^2 * e^( - 0.5u + 1 )
= u^2 * e * e^( - 0.5u )
= e * u^2 / e^(0.5u)
→ e * 2u / [ 0.5 * e^(0.5u) ] , 利用 the L'Hopital's rule
→ e * 2 / [ 0.25 * e^(0.5u) ] , L'Hopital's rule
= 8e / e^(0.5u)
→ 0

故此函數有水平漸進線 y = 0

Ans: y = 0


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