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Math Help - Need help with 2 limit-related calculus problems

  1. #1
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    Need help with 2 limit-related calculus problems

    Please don't just give me the answers (I can use my calculator for that), but explain how you get them by hand so I can understand them.

    1) limit as Δx approaches zero OF (cos(π+Δx) + 1) / Δx

    2) The function s(t) = -4.9t^2 + 200 gives the height in meters of an object that has fallen from a height of 200 meters. The velocity at time t = a seconds is given by: limit as t approaches a OF (s(a) - s(t)) / (a - t). At what velocity will the object impact the ground?

    I just can't seem to make the algebra work on these two! Thanks for your help!
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  2. #2
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    Quote Originally Posted by seuzy13 View Post
    1) limit as Δx approaches zero OF (cos(π+Δx) + 1) / Δx
    Expand \cos(\pi +\Delta x)=\cos(\pi)\cos(\Delta x)-\sin(\pi)\sin(\Delta x)
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  3. #3
    Super Member Aryth's Avatar
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    You have:

    s(t) = -4.9t^2 + 200

    First thing to do would be to evaluate the numerator of the limit:

    s(a) - s(t) = (-4.9a^2 + 200) - (-4.9t^2 + 200)

    = -4.9a^2 + 200 + 4.9t^2 - 200 = 4.9t^2 - 4.9a^2

    = -4.9(a^2 - t^2)

    I think you can get it from here.
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  4. #4
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    Quote Originally Posted by Plato View Post
    Expand \cos(\pi +\Delta x)=\cos(\pi)\cos(\Delta x)-\sin(\pi)\sin(\Delta x)
    I have done that, but don't know what to do next.
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  5. #5
    Super Member Aryth's Avatar
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    We know that:

    cos(\pi) = -1

    sin(\pi) = 0

    So,

    cos(\pi)cos(\Delta x) - sin(\pi)sin(\Delta x) = -cos(\Delta x)

    So, now you have:

    \frac{1 - cos(\Delta x)}{\Delta x}

    From here you're going to need to multiply both sides of the fraction by (1 + cos(\Delta x)), to get:

    \frac{1 - cos^2(\Delta x)}{\Delta x(1 + cos(\Delta x))}

    Now, the numerator looks pretty familiar and you can split the fractions:

    \lim_{\Delta x \to 0} \frac{sin(\Delta x)}{\Delta x} \lim_{\Delta x \to 0} \frac{sin(\Delta x)}{1 + cos(\Delta x)}

    From here you use the fact that we (should) know that:

    \lim_{x \to 0}\frac{sin(x)}{x} = 1

    Now, we have:

    (1)\lim_{\Delta x \to 0}\frac{sin(\Delta x)}{1 + cos(\Delta x)}

    I believe you can solve it from here.
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  6. #6
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    Another way to solve the first one:

    \lim_{\Delta x \to 0}\frac{cos(\pi + \Delta x) + 1}{\Delta x} = \lim_{\Delta x \to 0}\frac{cos(\pi + \Delta x) - cos(\pi)}{\Delta x} = -sin(\pi) = 0

    From the definition of derivatives (I'm not sure you were supposed to do it this way... have you proven that \frac{d}{dx}cos(x) = -sin(x) ? )
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  7. #7
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    I suspect that due to this time in a school year, this question comes in a chapter before differentiation is taught.
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  8. #8
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    Quote Originally Posted by Plato View Post
    I suspect that due to this time in a school year, this question comes in a chapter before differentiation is taught.
    Hmm, you're probably right... University only starts here mid-October so I was a bit confused :O
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  9. #9
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    Quote Originally Posted by Aryth View Post
    We know that:

    cos(\pi) = -1

    sin(\pi) = 0

    So,

    cos(\pi)cos(\Delta x) - sin(\pi)sin(\Delta x) = -cos(\Delta x)

    So, now you have:

    \frac{1 - cos(\Delta x)}{\Delta x}

    From here you're going to need to multiply both sides of the fraction by (1 + cos(\Delta x)), to get:

    \frac{1 - cos^2(\Delta x)}{\Delta x(1 + cos(\Delta x))}

    Now, the numerator looks pretty familiar and you can split the fractions:

    \lim_{\Delta x \to 0} \frac{sin(\Delta x)}{\Delta x} \lim_{\Delta x \to 0} \frac{sin(\Delta x)}{1 + cos(\Delta x)}

    From here you use the fact that we (should) know that:

    \lim_{x \to 0}\frac{sin(x)}{x} = 1

    Now, we have:

    (1)\lim_{\Delta x \to 0}\frac{sin(\Delta x)}{1 + cos(\Delta x)}

    I believe you can solve it from here.
    Yes, we haven't gotten to differentiation yet.

    However, couldn't I stop at this step:

    \frac{1 - cos(\Delta x)}{\Delta x}

    Because this limit is predefined as 0?

    Anyway, thank you so much. I was looking at it this whole time not realizing I could simply solve the sin(pi) and cos(pi). I was treating them like variables too! It's amazing how you can overlook something so simple!
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  10. #10
    Super Member Aryth's Avatar
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    You could if that's ok with your teacher. If you would like to solve it without using that particular definition, you can continue with the method provided. Haha.
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  11. #11
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    Quote Originally Posted by seuzy13 View Post
    Yes, we haven't gotten to differentiation yet.

    However, couldn't I stop at this step:

    \frac{1 - cos(\Delta x)}{\Delta x}

    Because this limit is predefined as 0?

    Anyway, thank you so much. I was looking at it this whole time not realizing I could simply solve the sin(pi) and cos(pi). I was treating them like variables too! It's amazing how you can overlook something so simple!

    I suspect that would not be fine, since:

    \lim_{\Delta x \to 0} \frac{ 1-cos(\Delta x)}{\Delta x} \to \frac{0}{0} which is an indeterminate form, and therefore you can not say that it is 0 without further investigation!
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  12. #12
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    Quote Originally Posted by Defunkt View Post
    I suspect that would not be fine, since:

    \lim_{\Delta x \to 0} \frac{ 1-cos(\Delta x)}{\Delta x} \to \frac{0}{0} which is an indeterminate form, and therefore you can not say that it is 0 without further investigation!
    Yes, but we were taught to just accept it as zero at that point, so it should be fine. Thanks.
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