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anonymous

  • 4 years ago

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

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  1. anonymous
    • 4 years ago
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    are you allowed to use l'hopital's rule?

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

  3. nenadmatematika
    • 4 years ago
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    1/2

  4. anonymous
    • 4 years ago
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    in any case the answer is \[\frac{1}{2}\]

  5. anonymous
    • 4 years ago
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    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

  6. anonymous
    • 4 years ago
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    i dont know the hospital rule that you have said :(

  7. anonymous
    • 4 years ago
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    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?

  8. nenadmatematika
    • 4 years ago
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    |dw:1328531016861:dw|

  9. nenadmatematika
    • 4 years ago
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    without using l'hopital rule

  10. anonymous
    • 4 years ago
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    any computation as a long the answer will be the same in a long method

  11. anonymous
    • 4 years ago
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    i dont know it sorry :(

  12. anonymous
    • 4 years ago
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    there is a nice neat answer above. i think you can also use \[\frac{sin(2x)\cos(4x)}{\sin(4x)}=\frac{\sin(2x)(\cos^2(2x)-\sin^2(2x))}{2\sin(2x)\cos(2x)}\]

  13. anonymous
    • 4 years ago
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    here i used the double angle formula for sine and cosine

  14. anonymous
    • 4 years ago
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    can i post the 3rd question too?

  15. anonymous
    • 4 years ago
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    then cancel the sin(2x) top and bottom, get \[\frac{\cos^2(2x)-\sin^2(2x)}{2\cos(2x)}\] replace x by zero, get \[\frac{1}{2}\]

  16. anonymous
    • 4 years ago
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    no limit to the amount you can post

  17. anonymous
    • 4 years ago
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    yes i will write your solution :)

  18. anonymous
    • 4 years ago
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    if f(x)=tanx-x and g(x)=x^3, evaluate the limit of f(x) over g(x) as x approaches 0. -3rd question.

  19. anonymous
    • 4 years ago
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    @nenadmatematica if you can do this without l'hopital or power series i will be impressed

  20. nenadmatematika
    • 4 years ago
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    haha I just wanted to ask you the same thing :D ....

  21. anonymous
    • 4 years ago
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    lol 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

  22. anonymous
    • 4 years ago
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    @nenadmatematika tahnks for helping us too :)

  23. nenadmatematika
    • 4 years ago
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    well 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

  24. anonymous
    • 4 years ago
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    i really dont know but if both of you wants to use L'hopital rule. then i will agree to both of you

  25. anonymous
    • 4 years ago
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    4th question: Evaluate the lim x^2-16 over x+4 as x->4.

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