Does the limit of 1/(lnx) as x goes to infinity equal zero?

Question

I feel like it doesn't because it decreases too slowly. I need it to find the value of the integral from 2 to infinity of 1/(x(lnx)squared). The integral ends up being -1/(lnb) + 1/(ln2). as b approaches infinity

Are you sure? Because I remember there being an example in class where it was getting smaller and smaller but it didn't imply convergence.

Answer 1

I think it actually depends on the direction of the curve of the function. I the curves move apart, then yes, the limit is infinitismal. If one curve overtakes the other, then the limit will evenually meet zero. You must know more than simply the integral of x. It does matter if it gets there slowly, with respect to proportions. If you halve x over and over, it will never truly reacy zero. If the curve is too gentle, it will will theoretically "straighten" before zero. But this can only occur in a single case.

Answer 2

Yes, that's correct. The limit of 1/lnx as x-->positive infinity is 0 . That's beacause lnx approaches positive infinity as x approaches positive infinity. So 1 divided by a very big number goes near zero as the divisor gets larger and larger and the whole thing (1/lnx) approaches 0 as x approaches positive infinity. Hence, the answer to the said integral is ln2.

Answer 3

yes, as x goes to infinity so does lnx...because it is 1/lnx, then the bigger the denominator the closer the number gets to zero..

think of it this way... 1/(x^2) you get answers like 1/4, 1/16, 1/36, 1/64....the higher that value you give to x the smaller the fraction becomes...so for your problem, it gets closer and closer to zero but never actually reaches it so the limit is indeed zero

Answer 4

Yes, its true.
Recall that lim (x ->inf) f(x) = 0 iff for every eps>0 there is a real number K>0 such that x>K implies |f(x)|< eps. Now to prove that lim (x->inf) 1/ln(x) = 0, take an arbitrary eps>0. We want to find K such that x > K implies 1/ln(x)

Answer 5

Yes. The limit of ln(x) --> ∞ as x--> ∞.

So the limit of 1/ln(x) --> 0 as x--> ∞, because 1/∞ = 0. It doesn't matter if it gets there slowly, only that it gets there.

Answer 6

JUST USE L'HOPITALS RULES;

limit x-------------------infinty

1/ln(x)(using l'hopitals rule find derivative of top and bottom)=

limit x---------infinty

0/(1/x)= limit x-------------infinty= 0(x/1)=0

therefore the answer is 0 as x approaches infinity

Answer 7

lim x^(a million/lnx) = lim e^ln( x^(a million/lnx)) on the grounds that e^[ln(A)]= A; that is a trick to make the calculation less demanding = lim e ^ [a million/ln(x) * ln(x)] utilising exponent regulation of logarithm : ln(A^B) = Bln(A) = lim e^a million = e

Answer 8

as x tends to infinity ln x tends to infinity and as ln x tends to infinity 1/ln x tends to zero