✔ 最佳答案
Question:
lim(x→0⁺) sinˣx
🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘
Soultion:
lim(x→0⁺) sinˣx
= lim(x→0⁺) (sin x)ˣ
= e^[ ln lim(x→0⁺) (sin x)ˣ ]
= e^[ lim(x→0⁺) x ln(sin x) ]
= e^{ lim(x→0⁺) ln(sin x) / (1/x) } ...... [ - ∞/∞ ]
= e^{ lim(x→0⁺) [(cos x)/(sin x)] / (-1/x²) } ...... [ Using L'Hôpital's rule ]
= e^[ - lim(x→0⁺) (x cos x) (x / sin x) ]
( As x→0⁺, lim (x / sin x) = 1 )
= e^[ - (0) (1) ]
= 1
🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘
L'Hôpital's rule ( 羅必達法則 )
If lim(x→c) f(x) = lim(x→c) g(x) = 0 or ±∞ and g'(x) ≠ 0,
then lim(x→c) f(x) / g(x) = L
🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘
20160707
相片:
我們發生某些問題,請再試一次。