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anonymous

  • 4 years ago

lim x^2cos2x/1-cosx x->0 Can someone explain to me how you solve this?

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  1. slaaibak
    • 4 years ago
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    Use L'Hospitals rule. If the limit is in an indeterminate form, the limit of f(x)/g(x) is equal to f'(x)/g'(x)

  2. anonymous
    • 4 years ago
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    I dont think I cant use that they'll mark me off on my test. We can only use calc I stuff.

  3. anonymous
    • 4 years ago
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    im suppose to use something along the lines of lim(h->0) sinh/h=1 or lim(h->0) cosh-1/h=0, I'm guessing because thats what they put in the examples.

  4. slaaibak
    • 4 years ago
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    You should be able to do partition it like that. I'll do it quickly

  5. anonymous
    • 4 years ago
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    the only rule i can see that applies is the quotient rule lim(x->c) f(x)/g(x)= lim f(x)/ lim g(x) if g(x) not equal to zero

  6. myininaya
    • 4 years ago
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    \[\frac{x^2 \cos(2x)}{1-\cos(x)}=\frac{x^2 \cos(2x)}{1-\cos(x)} \frac{1+cos(x)}{1+cos(x)}\] \[\frac{x^2 \cos(2x) (1+\cos(x))}{1-\cos^2(x)}\] \[\frac{x^2 \cos(2x)}{\sin^2(x)}=\frac{x^2}{\sin^2(x)} \cdot \cos(2x) =(\frac{x}{\sin(x)})^2 \cdot \cos(2x)\]

  7. myininaya
    • 4 years ago
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    that should help

  8. slaaibak
    • 4 years ago
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    \[\lim_{x \rightarrow 0} {x^2 \cos2x \over 1- \cos x} = \lim_{x \rightarrow 0} {x^2 \cos 2x (1+ cosx) \over \sin^2 x}\] Meh too late. What myinanaya said!

  9. myininaya
    • 4 years ago
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    oops i left over my (1+cos(x))

  10. myininaya
    • 4 years ago
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    sorry i'm on my slow laptop

  11. anonymous
    • 4 years ago
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    myininaya is the master of these trig limits

  12. anonymous
    • 4 years ago
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    sin^2x when x=0 is 0 that would make it undefined though

  13. anonymous
    • 4 years ago
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    or even with sinx on the bottom

  14. myininaya
    • 4 years ago
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    \[\lim_{x \rightarrow 0}\frac{x}{\sin(x)}=1\]

  15. slaaibak
    • 4 years ago
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    I didn't complete mine as myin was before me

  16. myininaya
    • 4 years ago
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    I just left off a factor by the way (1+cos(x)) so don't forget to look at that when you are doing this @ awstin

  17. anonymous
    • 4 years ago
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    makes you appreciate l'hopital

  18. anonymous
    • 4 years ago
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    I get it now thank you, I didnt know that x/sinx = 1 though lol

  19. anonymous
    • 4 years ago
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    l'hopital makes this easier?

  20. myininaya
    • 4 years ago
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    you actually prove that on the way to proving sin(x)/x->1 as x->0 when using squeeze thm

  21. anonymous
    • 4 years ago
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    cacl is hard :\ I was amazing at precalc too!

  22. anonymous
    • 4 years ago
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    calc*

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