I would like to know whether the limit as x tends to 0 of x/log x exists. If yes, then what is it ?
Thanks
cheers
niranjan

It's 0. You see, Log x -> -infinity , so 1/log x -> 0. Hence x/log x also tends to 0.

David

David,

Your argument is not valid.

Since Ln(x) -> -infinity
and x -> 0

you get, 0/-infinity

This is an indeterminant form

it is not possible to use l'Hopital's rule either.

How did you calculate this limit?

Brock Lynn
Bogalusa, Louisiana

It is not an indeterminate form. log x -> -infinity so (log x)^-1 -> 0. And x -> 0 also, so by the product of limits theorem, x (log x)^-1 also tends to zero.

Just to be annoying, I think the limit Niranjan took ought to be x tends to 0⁺ . If x tends to 0 from below, then we have to specify which branch of log we are taking. This actually won't matter much so I'll assume we do this on the complex plane (you can restrict the domain later), and it is obvious that whatever branch cut you made, |log z|> = log |z| (since log z = log |z| + i arg(z)).

For |x|< 1/e, x non-zero, we have |log x|> 1, so we have

0< |x/log x|=|x|/|log x|< |x|/1=|x|

and we have 'sandwiched' |x/log x| between 0 and something that tends to 0 as x tends to 0. Hence |x/log x| tends to 0 as x tends to 0, which is exactly the statement x/log x tends to0 as x tends to 0.

Kerwin