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Math Help - Need help with evaluating Trig Limits

  1. #1
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    Question Need help with evaluating Trig Limits

    Hey, this is my first time posting here so I will try to keep this brief!

    So I am redoing pre-calculus to upgrade my math for post-secondary, and it's been a year since I did this in grade 12 so I'm pretty rusty. =(

    I am going through limits at the moment and I came across the equation,

    lim sinAx = A/B
    x→0 Bx

    Or otherwise

    lim sinx = 1
    x→0 x

    Now this has helped me greatly in evaluating limits, but I remember there being a cosine limit something like this that also equals 1 which I could use to isolate and simplify the problem, but I just can't remember and I can't seem to find it in my textbooks, does anyone know of anything like this?

    Thanks you very much for the help!
    Kasper
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  2. #2
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    Quote Originally Posted by Kasper View Post
    Hey, this is my first time posting here so I will try to keep this brief!

    So I am redoing pre-calculus to upgrade my math for post-secondary, and it's been a year since I did this in grade 12 so I'm pretty rusty. =(

    I am going through limits at the moment and I came across the equation,

    lim sinAx = A/B
    x→0 Bx

    Or otherwise

    lim sinx = 1
    x→0 x

    Now this has helped me greatly in evaluating limits, but I remember there being a cosine limit something like this that also equals 1 which I could use to isolate and simplify the problem, but I just can't remember and I can't seem to find it in my textbooks, does anyone know of anything like this?

    Thanks you very much for the help!
    Kasper
    i'n not sure if there is one for cosine, but it is true that:

    \lim_{x \rightarrow 0} \frac{x}{\sin{x}}=1
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  3. #3
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    Quote Originally Posted by GaloisTheory1 View Post
    i'n not sure if there is one for cosine, but it is true that:

    \lim_{x \rightarrow 0} \frac{x}{\sin{x}}=1

    further thinking about this made me realize that there is not one with cosine

    the reason is that in calculus, you will/have learned that

    \lim_{x \rightarrow 0} \frac{x}{\sin{x}}=1 is in an indeterminant form so you can use l'hopital's rule to get the limit.

    however,

    \lim_{x \rightarrow 0} \frac{x}{\cos{x}}=1

    is not in an indeterminant form. if this doesn't make sense now, just wait until you take calculus.
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  4. #4
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    I believe you are thinking of
    \lim_{x\rightarrow 0} \frac{1- cos(x)}{x}= 0
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    I believe you are thinking of
    \lim_{x\rightarrow 0} \frac{1- cos(x)}{x}= 0
    That's it! Thanks alot, that's what I was looking for.
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