Question

lim x->0^+ (sinx*lnx) can someone explain to me how this is zero

I put ln x/ 1/sinx because usually sinx and x would be around the same number if its right by zero giving me -inf/inf but I'm not really sure what to do after it doesn't seem like thats proper

Answer 1

you are in the form \(0\times \infty\) so standard trick it to write it as a fraction as
\[\frac{\sin(x)}{\frac{1}{\ln(x)}}\] or
\[\frac{\ln(x)}{\csc(x)}\] which ever you think will work better
probably the second one

Answer 2

even after taking the derivative I get cosx/ xsinx+cosx with means something is def wrong

Answer 3

ahh kk

Answer 4

so cosx*x=0? thats what we end up with?

Answer 5

derivative of \(\csc(x)\) is \(-\csc(x)\cot(x)\)

Answer 6

i did it with sinx/ 1/lnx

Answer 7

wait cosx*x isn't right then

Answer 8

i wouldn't use that one since derivative of \(\frac{1}{\ln(x)}\) is a worse

Answer 9

oh well fml i thought for a sec that it was x lol didn't look at that properly

Answer 10

holy crap this is a huge pain

Answer 11

after the algebra for the one i wrote you should get
\[-\frac{\sin^2(x)}{x\cos(x)}\] if i did the algebra right. then you have \(\frac{0}{0}\) so one more time should do it

Answer 12

oh wow I should've just kept it 1/sinx then

Answer 13

muahaha! got it thanks!

Answer 14

yes it amounts to the same thing. the derivative of
\[\frac{1}{\sin(x)}\] is \[-\frac{\cos(x)}{\sin^2(x)}\] then flip it

Answer 15

stupid arithmetic errors

Answer 16

i like to think of them as "typos"

Answer 17

haha I've just been making so many of them because of lack of sleep.... my test is in 3 hours and I've been studying till 2am every night and waking up at 8am