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Mathematics 7 Online
OpenStudy (anonymous):

2nd question: Find lim (sin2xcot4x) X->0.

OpenStudy (anonymous):

are you allowed to use l'hopital's rule?

OpenStudy (anonymous):

assuming this is \[\lim_{x\rightarrow 0}\sin(2x)\cot(4x)\] \[\lim_{x\rightarrow 0}\frac{\sin(2x)\cos(4x)}{\sin(4x)}\]

OpenStudy (nenadmatematika):

1/2

OpenStudy (anonymous):

in any case the answer is \[\frac{1}{2}\]

OpenStudy (anonymous):

but if you need to show your work your answer will depend on what you are allowed to use. l'hopital's rule is the simplest, otherwise it will be some work

OpenStudy (anonymous):

i dont know the hospital rule that you have said :(

OpenStudy (anonymous):

you mean you do not know it or you are not allowed to use it? i assume this is calc class, so hve you covered deriviatives yet or are you just starting out?

OpenStudy (nenadmatematika):

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