lim x->- infinity (e^(x-2)/(x-1)) with l'hopital's rule is 0 Can someone explain why is 0 ?
回答 (1)
2016-08-24 2:42 pm
lim x-->-∞ e^(x-2) /(x-1)
You cannot apply L'Hopital's rule because (x-1) is not in the exponent.
= lim x-->-∞ e^(x-2) / lim x-->-∞ (x-1)
= lim x-->-∞ e^x / (e^2 lim x-->-∞ (x-1))
As x approaches -∞, the numerator approaches 0 while the denominator approaches-∞. Therefore, the ratio approaches 0.
You cannot apply L'Hopital's rule because (x-1) is not in the exponent.
= lim x-->-∞ e^(x-2) / lim x-->-∞ (x-1)
= lim x-->-∞ e^x / (e^2 lim x-->-∞ (x-1))
As x approaches -∞, the numerator approaches 0 while the denominator approaches-∞. Therefore, the ratio approaches 0.
收錄日期: 2021-04-21 19:42:29
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