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Thread: Limit help (two problems)

  1. #1
    Newbie riddikulus's Avatar
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    Limit help (two problems)

    I have been trying to figure out these limit problems and I get stuck on trying to move the function around to make it possible to solve.

    The first one is:

    lim sinx - cosx / tanx - 1
    x--> (pi/4)

    All I've gotten so far is to replace tanx with sinx/cosx but I don't even know if that's the way to go...


    the second is:

    lim sqrt(a + 2h) - sqrt(a) / h
    h-->0

    I tried multiplying by the conjugate, but when I multiply things out, I get stuck at:

    2h / h*sqrt(a+2h) + sqrt(a)

    pleaseeee help me. I have no idea how to move on from here; I'm terrible at figuring out what to do next in these situations.
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  2. #2
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    Re: Limit help (two problems)

    Quote Originally Posted by riddikulus View Post
    The first one is:
    lim sinx - cosx / tanx - 1
    x--> (pi/4)
    $\displaystyle \frac{\sin(x)-\cos(x)}{\tan(x)-1}=\frac{\cos(x)[\sin(x)-\cos(x)]}{\sin(x)-\cos(x)}$
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  3. #3
    Newbie riddikulus's Avatar
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    Re: Limit help (two problems)

    where did you get that though?
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    Re: Limit help (two problems)

    Quote Originally Posted by riddikulus View Post
    where did you get that though?
    Multiply numerator and denominator by $\displaystyle \cos(x).$
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  5. #5
    Newbie riddikulus's Avatar
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    Re: Limit help (two problems)

    okay but then you still get zero on the bottom, so what do you do next? i seriously have no idea how to shift trig fxns or anything, for that matter.
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    Re: Limit help (two problems)

    Quote Originally Posted by riddikulus View Post
    okay but then you still get zero on the bottom, so what do you do next? i seriously have no idea how to shift trig fxns or anything, for that matter.
    Can you do basic algebra?
    Divide out $\displaystyle \sin(x)-\cos(x)$
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