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Jusaquikie
Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim sin x ln 4x x→0+
I tried re-writing it as \[\frac{ \ln 4x }{ \csc x }\] but the derivative of csc is always going to have csc in it and csc 0 is 1/sin 0 which is 1/0 no matter how many derivatives i take
well the answer is zero, but i'm still not sure why, other than it's not zero but a really small number. thanks for the help.