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anonymous
 4 years ago
2nd question: Find lim (sin2xcot4x) X>0.
anonymous
 4 years ago
2nd question: Find lim (sin2xcot4x) X>0.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0are you allowed to use l'hopital's rule?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0assuming this is \[\lim_{x\rightarrow 0}\sin(2x)\cot(4x)\] \[\lim_{x\rightarrow 0}\frac{\sin(2x)\cos(4x)}{\sin(4x)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in any case the answer is \[\frac{1}{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i dont know the hospital rule that you have said :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you 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?

nenadmatematika
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1328531016861:dw

nenadmatematika
 4 years ago
Best ResponseYou've already chosen the best response.1without using l'hopital rule

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0any computation as a long the answer will be the same in a long method

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i dont know it sorry :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there is a nice neat answer above. i think you can also use \[\frac{sin(2x)\cos(4x)}{\sin(4x)}=\frac{\sin(2x)(\cos^2(2x)\sin^2(2x))}{2\sin(2x)\cos(2x)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0here i used the double angle formula for sine and cosine

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can i post the 3rd question too?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then cancel the sin(2x) top and bottom, get \[\frac{\cos^2(2x)\sin^2(2x)}{2\cos(2x)}\] replace x by zero, get \[\frac{1}{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no limit to the amount you can post

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes i will write your solution :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if f(x)=tanxx and g(x)=x^3, evaluate the limit of f(x) over g(x) as x approaches 0. 3rd question.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@nenadmatematica if you can do this without l'hopital or power series i will be impressed

nenadmatematika
 4 years ago
Best ResponseYou've already chosen the best response.1haha I just wanted to ask you the same thing :D ....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol well, i guess we cannot give an elementary reason for this. the answer is \[\frac{1}{3}\] but i cannot think of a gimmick to simplify this expression. are you sure you have not covered l'hopital? because i am stumped. in particular you have a trig fuction combined in combination with x and x^3 so there is no simple trig identity that will change the form of this for you

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@nenadmatematika tahnks for helping us too :)

nenadmatematika
 4 years ago
Best ResponseYou've already chosen the best response.1well you're welcome....I agree with satellite that this example is very convenient for using L'Hopital rule.....I can't think of any other way now :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i really dont know but if both of you wants to use L'hopital rule. then i will agree to both of you

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.04th question: Evaluate the lim x^216 over x+4 as x>4.
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