Question

Is this right? (I'm using L'H rule)

lim ln(x)sin(x) = ln(x)/1/sin(x)
x->0^+ L'H

= 1/x (1(cos(x))/ sin^(2)(x)) = sin^(2)(x)/xcos(x)

= lim(x) tan(x)sin(x)/x
x->0^(+)

= lim (sec^(2)(x)sin(x) + tan(x)cos(x))/1
L'H x->0^(+)

= 0

Answer 1

Note I was lazy with limit notation :L

Answer 2

\[\lim_{x \rightarrow 0^+} \ln(x)\sin(x)= \lim_{x \rightarrow 0^+} \frac{\ln(x)}{\csc(x)}=\lim_{x \rightarrow 0^+} \frac{\frac{1}{x}}{-\csc(x)\cot(x)}=\lim_{x \rightarrow 0^+}\frac{-\sin^2(x)}{x \cos(x)}\]

You are again at a 0/0 so you need to use it again...

\[\lim_{x \rightarrow 0^+}\frac{-(\cos(x)\tan(x)+\sin(x)\sec^2(x))}{1}=0\]
Sorry for typing it, i work it out as I type it.