limit x^4e^3x
x-> - infinity
at the step where i applied L'rule for the 3rd time, how come i didn't get infinity/ infinity . ( 24x) / ( -27e^-3x) would be inf/ -inf...
plz clear up my question. i know how to get the problem itz jez 1 step i didnt clearly get.
2007-04-30 21:30:41 · 2 answers · asked by ruby 2 in Science & Mathematics ➔ Mathematics
ok i got the 1st question i asked
2nd question:
the last step...(24) / (81e^-3x) how is it 0?
does a big # over a lil # equal 0?
2007-04-30 21:45:34 · update #1
Actually 24x is like -∞ too, so it's really -∞/-∞. But the sign doesn't matter; if they're both going to -∞, or one is going to ∞ and the other to -∞, we still call this an ∞/∞ form.
2007-04-30 21:35:35 · answer #1 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
You apply L'Hospital 4 times to get
lim (x-> -inf) 24/81e^(-3x).
When x goes to -inf, -3x goes to positive inf and therefore e^(-3x) also goes to inf, as the numerator of the fraction is a finite number, you get a quotient which is very small and hence the limit is zero.
2007-04-30 22:22:43 · answer #2 · answered by firat c 4 · 0⤊ 0⤋