lim x->inf ln(x)/x^p=0
2006-10-04 09:23:39 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
limx->inf ln|x|/x^p = inf/inf^p --> INDETERMINANT
using L'Hopital's Rule,
limx->inf ln|x|/x^p = limx->inf[(1/x)/(px^(p-1)0]
=limx->inf (1/x)(1/px^[p-1])
=limx->inf 1/px^p = 0
therefore TRUE
2006-10-04 09:38:14 · answer #1 · answered by hackmaster_sk 3 · 0⤊ 0⤋
As x->inf, ln(x) -> inf, and x^p -> inf for p>0.
Therefore, you can use l'Hospital's Rule and get
lim x->inf (1/x) / px^(p-1)
= lim x->inf 1/(px^p) = 0
So the statement is true, if p>0.
2006-10-04 16:26:40 · answer #2 · answered by James L 5 · 0⤊ 0⤋
Indeterminate form of inf./inf. Apply L'Hopital's rule.
2006-10-04 16:32:55 · answer #3 · answered by Anonymous · 0⤊ 0⤋
what The heck is that, dude i got 2 times 100 on my marking period then 98 and 95 on math in 9th had the highest on the regents but i don't get what the heck is that
2006-10-04 16:29:07 · answer #4 · answered by cool dude 2 · 0⤊ 0⤋
I forgot calculas, but it seems true!
Growth rate of Log(x) is much less than x^p
2006-10-04 16:36:02 · answer #5 · answered by Arash Salimi 1 · 0⤊ 0⤋
I forget. When all else fails flip a coin.
2006-10-04 16:27:46 · answer #6 · answered by Grev 4 · 0⤊ 0⤋
guess
2006-10-04 16:31:14 · answer #7 · answered by guess 3 · 0⤊ 0⤋