# Thread: Calculation of a limit using L'Hopital

1. ## Calculation of a limit using L'Hopital

Hello

I'm having trouble calculating the right hand limit of sqrt(x) * ln(x) at 0

I see that it's an indeterminate form of the type 0 * -infinity
So I tried turning into the limit of sqrt(x) / (1/lnx) so that it would be a case of 0/0
and applied L'Hopitals rule, but I got that it's equal to
1/2 * limit at 0+ of sqrt(x) * (lnx)^2

In other words back where I started...what am I doing wrong?

2. Originally Posted by moses
Hello

I'm having trouble calculating the right hand limit of sqrt(x) * ln(x) at 0

I see that it's an indeterminate form of the type 0 * -infinity
So I tried turning into the limit of sqrt(x) / (1/lnx) so that it would be a case of 0/0
and applied L'Hopitals rule, but I got that it's equal to
1/2 * limit at 0+ of sqrt(x) * (lnx)^2

In other words back where I started...what am I doing wrong?
Try doing it the other way.
$\sqrt{x}ln(x) = \frac{ln(x)}{\frac{1}{\sqrt{x}}}$

-Dan

3. Oh okay that works, thanks