thank you very much but i didn't mean the derivative what i meant was to prove that the limit as x approaches zero to the natural log of x that the limit will approach - inf. by the epsilon-delta definition a limit

I'm a little rusty but I came up with this so there are no guarantees on the logic. If some out there can correct me I'd be grateful. It sounds O.K. to me but one is never quite sure with these.

First note that we have to approach 0 from the right since Domain of ln(x) = set of positive real numbers. Also note that for 0 < x < 1 that ln(x) is negative.

We must show that for any d between 0 an 1 and 0 < x < d there must exist a large enough N >0 such that ln(x) < -N.

i.e. ln(x) decreases without bound. More formally:

If 0 < x < d then we can find an N > 0 so that ln(x) < -N

So choose d = 1/e^N (which is between 0 and 1)

So now we have (from above):

x < d

x < 1/e^N

ln(x) < ln(1/e^N)

ln(x) < -N since ln(1/e^N) = -N

In other words, no matter how small we choose d, we can always find a large enough N>0 so that ln(x) < -N

So now ln(x) is lower than -N and decreases without bound.

thank you very much but i didn't mean the derivative what i meant was to prove that the limit as x approaches zero to the natural log of x that the limit will approach - inf. by the epsilon-delta definition a limit

I'm a little rusty but I came up with this so there are no guarantees on the logic. If some out there can correct me I'd be grateful. It sounds O.K. to me but one is never quite sure with these.

First note that we have to approach 0 from the right since Domain of ln(x) = set of positive real numbers. Also note that for 0 < x < 1 that ln(x) is negative.

We must show that for any d between 0 an 1 and 0 < x < d there must exist a large enough N >0 such that ln(x) < -N.

i.e. ln(x) decreases without bound. More formally:

If 0 < x < d then we can find an N > 0 so that ln(x) < -NSo choosed = 1/e^N(which is between 0 and 1)So now we have (from above):x < dx < 1/e^Nln(x) < ln(1/e^N)ln(x) < -N since ln(1/e^N) = -NIn other words, no matter how small we choose d, we can always find a large enough N>0 so that ln(x) < -N

So now ln(x) is lower than -N and decreases without bound.