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anonymous

  • 5 years ago

find the limits when h ->0 h /tan^2h (calculus 1 style)

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  1. anonymous
    • 5 years ago
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    I think Lokisan did this for you last night. What don't you understand?

  2. anonymous
    • 5 years ago
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    because i need to know on the trig way what she told me i understood very clear

  3. anonymous
    • 5 years ago
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    but i need to do it without the l'hopitals way

  4. anonymous
    • 5 years ago
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    and i dont know :(

  5. anonymous
    • 5 years ago
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    \[\lim_{h \rightarrow 0}\frac{h.\cos^2h}{\sin^2h} = \frac{0}{0} = UND\] you have to use L'hopital's rule since you have 0/0 which is an undertermined form.

  6. anonymous
    • 5 years ago
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    ok,, but the professor told me i have to trig function that can help me to define this sinh /h and cosh-1

  7. anonymous
    • 5 years ago
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    but why make it so complicated lol? :)

  8. anonymous
    • 5 years ago
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    mm calculus 1... i am not in cal 2... they didnt teach me the l'hopital way

  9. anonymous
    • 5 years ago
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    :(

  10. anonymous
    • 5 years ago
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    lol, this is calculus I too :) believe me

  11. anonymous
    • 5 years ago
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    l'hopital is one of the first things taught in Cal I

  12. anonymous
    • 5 years ago
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    if you understand L'hopital's rule, then you won't find difficulty solving such problems

  13. anonymous
    • 5 years ago
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    mm.. so can u teachme de difficult way

  14. anonymous
    • 5 years ago
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    i understand it now... but he dont want to look at that in that way

  15. anonymous
    • 5 years ago
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    LOL, I'm afraid my profs skipped the difficult way ^_^" and I'm only aware of this way and many others too

  16. anonymous
    • 5 years ago
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    jajaajaja,,,

  17. anonymous
    • 5 years ago
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    lol, they skipped the cosh and sinh from our syllabus

  18. anonymous
    • 5 years ago
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    :(....

  19. anonymous
    • 5 years ago
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    OK, here is the other method (not necessarily recommended for this problem) h lim (h cos h)/h multiplied by lim (h cos h/h) divided by lim (h tan h)/h (htan/h)

  20. anonymous
    • 5 years ago
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    why cosh multiplied by cosh

  21. anonymous
    • 5 years ago
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    I am just expanding cos^2 h. You can take the lim of one cos h and multiply by another cos h

  22. anonymous
    • 5 years ago
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    ok so, if we change h/tan^2h by h/sin^2h/cos^h

  23. anonymous
    • 5 years ago
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    Look this has too many terms to understand this method. This method is usually done for something simple like lim (sin x). If you wanted to us that method on this simple one you multiply by x and divide by x. (You are essentially dividing by 1; it changes nothing.) You are trying to manipulate it because sin x/x =1.

  24. anonymous
    • 5 years ago
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    ohhhhh

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