Question

How do you solve:
lim x->0+ sin(x)ln(2x)

Answer 1

if you replace x by 0 you get
\[0\times -\infty \] right?

Answer 2

right

Answer 3

so standard l'hopital trick is to rewrite at either
\[\frac{\ln(2x)}{\csc(x)}\] or
\[\frac{\sin(x)}{\frac{1}{\ln(2x)}}\]

Answer 4

Then choose the one where you get 0/0 or infinity/infinity

Answer 5

that's really helpful! thank you, i know what to do from there

Answer 6

i think you have to use l'hopital twice. at least it did.

Answer 7

yeah, I did.

also, if i have a question that says :
\[\ln ((1-9x)^{1/x})= (blank)(\ln(1-9x))\]

what am I suppose to do?