can i apply l'hospital's rule to negative infinity minus infinty? ex. lim f(x)-g(x)= -(inf)-(inf)
2007-01-27 12:10:32 · 3 answers · asked by joe s 1 in Science & Mathematics ➔ Mathematics
need more answers please
2007-01-27 12:20:40 · update #1
No, but sometimes you can use some algebraic manipulation on f(x) and g(x) to get to the indeterminate for \infty/\infty or 0/0.
2007-01-27 12:16:55 · answer #1 · answered by goldenflaws 2 · 0⤊ 0⤋
No, not directly. For sums and differences of indeterminate form, you may have to get creative.
Fact is, though, you don't need to do any such thing, if I've read you right. If f(x) goes to minus infinity, and g(x) to positive infinity (sending -g(x) to negative infinity), then f(x) - g(x) goes to minus infinity, so no indeterminate form.
2007-01-27 20:17:07 · answer #2 · answered by Anonymous · 0⤊ 0⤋
as you write it -(inf)-(inf)gives -(inf) There is no indetermination
If you have an indetermination inf -inf you CAN NOT apply
L´Hôpital directly You must write it as
lim[f(x)-g(x)] = lim [1/g(x) -1/f(x)]/ [1/f(x) *1/g(x)] which is indeterminate 0/0
2007-01-27 20:28:13 · answer #3 · answered by santmann2002 7 · 1⤊ 0⤋