Jusaquikie
  • 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+
Mathematics
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Jusaquikie
  • 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+
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Jusaquikie
  • Jusaquikie
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
anonymous
  • anonymous
|dw:1352704969770:dw|
anonymous
  • anonymous
|dw:1352705042857:dw|

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Jusaquikie
  • Jusaquikie
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.

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