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Math Help - Help with two limit problems...

  1. #1
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    Help with two limit problems...

    What is the limit of tanx/x as x approaches zero?
    and
    What is the limit of sin2x/sin3x as x approaches zero?
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  2. #2
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    l'hopital's rule

    Have you learned L'hopital's rule for limits yet?
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  3. #3
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    No, I haven't. Is it necessary to do these problems?
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  4. #4
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    no

    Well, no. It would just be easier since it is in indeterminate form.
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    L'Hopital's rule is not necessary
    Quote Originally Posted by MegaVortex7 View Post
    What is the limit of tanx/x as x approaches zero?
    hint: \frac {\tan x}x = \frac {\sin x}{x \cos x} = \frac 1{\cos x} \cdot \frac {\sin x}x

    and you should be able to handle that. think "special limit"

    What is the limit of sin2x/sin3x as x approaches zero?
    use the addition formula for sine to simplify. you may have to do it more than once

    \sin (A + B) = \sin A \cos B + \sin B \cos A

    in particular, \sin 2A = 2 \sin A \cos A and \sin 3A = \sin (2A + A)
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  6. #6
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    Thanks! However..I'm still a little unsure as to what to do for the second one. How would I go about expanding those?
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  7. #7
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    ok so

    [lim x->0] sin2x/sin3x=
    [lim x->0] 2sinxcosx/(4sinxcos^2(x) -sinx)=
    [lim x->0] 2cosx/(4cos^2(x)-1)
    = 2/3

    sin3x=
    sin (2x+x)=
    sin2xcosx + sinxcos2x=
    2sinxcos^2(x)+ sinx(2cos^2(x)-1)=
    4sinxcos^2(x) - sinx
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  8. #8
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    OH ok thank you very much! I still have some other questions though (on different problems...)

    WHat is
    [lim x->3 from the right] (sqrt(t^2-9))/(t-3)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by MegaVortex7 View Post
    OH ok thank you very much! I still have some other questions though (on different problems...)

    WHat is
    [lim x->3 from the right] (sqrt(t^2-9))/(t-3)
    post new questions in a new thread and recall that t^2 - 9 = (t + 3)(t - 3)

    (of course, you already posted this question here, so don't bother making a new thread for this. (it has happened before, so i just thought i'd say it))
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  10. #10
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    Alright, thanks. I just figured that posting it in the same thread would be less spam-y?
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by MegaVortex7 View Post
    Alright, thanks. I just figured that posting it in the same thread would be less spam-y?
    haha, ok. well, it's not.

    did you get the limit?
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  12. #12
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    I didn't. =/ If I break it into (t-3)(t+3), do I have to square it to get rid of the square root? What do I do from there?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by MegaVortex7 View Post
    I didn't. =/ If I break it into (t-3)(t+3), do I have to square it to get rid of the square root? What do I do from there?
    you cannot just square stuff. you can possibly change the value of the function and hence its limit. cancel where you can cancel and see what happens. you should notice there is a common (t - 3) in the numerator and denominator
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  14. #14
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    One is under a radical sign though. That's where I'm getting stuck. The entire numerator is a square root
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    Quote Originally Posted by MegaVortex7 View Post
    One is under a radical sign though. That's where I'm getting stuck. The entire numerator is a square root
    the square root is represented by the power 1/2. you can distribute the power. then use the rule \frac {x^a}{x^b} = x^{a - b}
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