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anonymous

  • 4 years ago

limit x->0, tan2x/x find the limit, help!!

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  1. anonymous
    • 4 years ago
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    l'Hopital's rule top and bottom. (2sec(x)^2)/1, limx->0 of 2sec(x)^2

  2. anonymous
    • 4 years ago
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    we havent learned that rule, yet, so u wouldnt have tan=cos/sin?

  3. anonymous
    • 4 years ago
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    sin/cos*

  4. TuringTest
    • 4 years ago
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    is it\[\tan(2x)\]or\[\tan^2(x)\]?

  5. anonymous
    • 4 years ago
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    tan2x not ^2

  6. anonymous
    • 4 years ago
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    I wonder, I don't think it's possible to solve tan(2x)/x without l'Hopital's. Unless I'm derping. Maybe some weird half angle identity?

  7. anonymous
    • 4 years ago
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    my calc class just started last week n my professor has mentioned that rule but has yet to teach it, so wa does that rule state?

  8. anonymous
    • 4 years ago
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    http://mathworld.wolfram.com/LHospitalsRule.html Apparently I spelled it wrong, but w/e. The proof is lovely.

  9. anonymous
    • 4 years ago
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    you could try \[\lim_{x\to 0}\frac{\sin(2x)}{x\cos(2x)}\]

  10. anonymous
    • 4 years ago
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    thats as far as i got what do i after that satel.?

  11. anonymous
    • 4 years ago
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    break into to parts \[\lim_{x\to 0}\frac{1}{\cos(2x)}= 1\] and \[\lim_{x\to 0}\frac{\sin(2x)}{x}=2\]

  12. TuringTest
    • 4 years ago
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    How about\[\frac{\tan(2x)}x=\frac{\sin(2x)}{x\cos(2x)}=2\frac{\sin x}{x}\cdot\frac{\cos x}{2\cos^2x-1}=2\]sat beat me as usual...

  13. anonymous
    • 4 years ago
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    yeah but you are quicker because i did not explain my second limit

  14. anonymous
    • 4 years ago
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    Man, algebra. I forget that if I can't solve a problem, it's probably just me doing algebra badly.

  15. Zarkon
    • 4 years ago
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    I would write \[\frac{\sin(2x)}{x}=2\frac{\sin(2x)}{2x}\]

  16. TuringTest
    • 4 years ago
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    dang...lol

  17. anonymous
    • 4 years ago
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    or replace x by x/2

  18. anonymous
    • 4 years ago
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    turning test, so 2cos^2x-1 cancels the top, n ur left wit 2?

  19. TuringTest
    • 4 years ago
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    I didn't even need to do that identity, don't know why I did but you can just plug in x=0 to that second part and get 1

  20. anonymous
    • 4 years ago
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    the answer is 2 tho

  21. TuringTest
    • 4 years ago
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    that's what you get in total\[\frac{\tan(2x)}x=\frac{\sin(2x)}{x\cos(2x)}=2\frac{\sin x}{x}\cdot\frac{\cos x}{\cos(2x)}=2\cdot1\cdot1=2\]

  22. TuringTest
    • 4 years ago
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    I didn't write the limits in ...lazy I know

  23. anonymous
    • 4 years ago
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    got it!!

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