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Math Help - Calculas- limits

  1. #1
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    Post Calculas- limits

    Apply limit laws to evaluate or show that the limit does not exist.

    lim Cos (π + ∆ x) +1
    ∆x -> 0 ∆x


    I started like this but It didnt go anywhere:

    lim Cos (π + ∆ x) +1 . Cos (π + ∆ x) -1 = lim Cos (π + ∆ x)^2 -1
    ∆x -> 0 ∆x Cos (π + ∆ x) -1 ∆x -> 0 ∆x(Cos(π + ∆ x) -1

    please help me thanks[IMG]file:///C:/DOCUME%7E1/Shayan/LOCALS%7E1/Temp/moz-screenshot-5.jpg[/IMG][IMG]file:///C:/DOCUME%7E1/Shayan/LOCALS%7E1/Temp/moz-screenshot-6.jpg[/IMG]
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  2. #2
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    Hi, meme22!

    Quote Originally Posted by meme22 View Post
    Apply limit laws to evaluate or show that the limit does not exist.

    lim Cos (π + ∆ x) +1
    ∆x -> 0 ∆x


    I started like this but It didnt go anywhere:

    lim Cos (π + ∆ x) +1 . Cos (π + ∆ x) -1 = lim Cos (π + ∆ x)^2 -1
    ∆x -> 0 ∆x Cos (π + ∆ x) -1 ∆x -> 0 ∆x(Cos(π + ∆ x) -1
    I assume this is the problem you are trying to solve:

    \text{Evaluate }\lim_{\Delta x\to0}\frac{\cos\left(\pi + \Delta x\right) + 1}{\Delta x}\text{.}

    Is that right?

    First of all, you can easily see what this limit should be if you observe that

    \lim_{\Delta x\to0}\frac{\cos\left(\pi + \Delta x\right) + 1}{\Delta x} = \lim_{\Delta x\to0}\frac{\cos\left(\pi + \Delta x\right) - (-1)}{\Delta x}

     = \lim_{\Delta x\to0}\frac{\cos\left(\pi + \Delta x\right) - \cos\left(\pi\right)}{\Delta x} = \cos'\left(\pi\right) = -\sin\left(\pi\right) = 0.

    So, we should suspect the limit to be zero. Now, to solve it using the properties of limits, here are some hints:

    1. Use the fact that \cos\left(u\pm v\right) = \cos u\cos v\mp\sin u\sin v to rewrite the numerator

    2. Make sure that you are familiar with this special limit:

    \lim_{x\to0}\frac{\cos\left(x\right) - 1}x = 0

    Does that help?
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  3. #3
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    Yes, thank you I figured it out
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