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Math Help - [SOLVED] formal computation limits

  1. #1
    Member sinewave85's Avatar
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    [SOLVED] formal computation limits

    I am stuck on these problems, taking the limit at x = 2.

     f(x) = \frac {sin(\pi x)} {x-2}

     f(x) = \frac {cos \frac{\pi} {x}} {x-2}

     f(x) = \frac {\frac{1}{x} -1} {x-2}

    Any help is greatly appreciated, including general advice about working these kinds of problems. What do you do when you can't "see" how to rationalize an equation?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by sinewave85 View Post
    I am stuck on these problems, taking the limit at x = 2.

     f(x) = \frac {sin(\pi x)} {x-2}

     f(x) = \frac {cos \frac{\pi} {x}} {x-2}
    Think about the definition of the derivative : g'(a)=\lim_{x\to a}\frac{g(x)-g(a)}{x-a}

    For example, for the first limit : \frac {\sin(x\pi)} {x-2}=\frac {\sin( x\pi)-\sin(2\pi)} {x-2}\underset{x\to2}{\to} ?
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  3. #3
    Member sinewave85's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

    Think about the definition of the derivative : g'(a)=\lim_{x\to a}\frac{g(x)-g(a)}{x-a}

    For example, for the first limit : \frac {\sin(x\pi)} {x-2}=\frac {\sin( x\pi)-\sin(2\pi)} {x-2}\underset{x\to2}{\to} ?

    I'm sorry, i'm at the very beginning of the course and they have not introduced the concept of the derivative yet. (I think that is in the next lesson.) I'm supposed to find the limit (if there is one) by using fractional reduction or by rationalizing the expression (or show that it can't be done).
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  4. #4
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    Hello,

    For the first one, substitute : t=x-2.

    When x \to 2, t \to 0.
    And x=t+2


    ---> L=\lim_{x \to 2} \frac{\sin(\pi x)}{x-2}=\lim_{t \to 0} \frac{\sin(\pi (t+2)}{t}

    L=\lim_{t \to 0} \frac{\sin(\pi t+2 \pi)}{t}

    But we know that for all x, \sin(x+2 \pi)=\sin(x)

    Therefore L=\lim_{t \to 0} \frac{\sin(\pi t)}{t}

    Substitute again, u=\pi t ---> \frac 1t=\frac{\pi}{u}


    L=\lim_{u \to 0} \pi \frac{\sin(u)}{u}

    This is supposed to be a well-known limit :/

    -----------------

    If you've learnt l'H˘pital's rule, you can use it in the first step
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    Quote Originally Posted by sinewave85 View Post
     f(x) = \frac {sin(\pi x)} {x-2}
    Using the limit composite rule this is equivalent to,
    \lim_{x\to 0} \frac{\sin [\pi(x+2)]}{x} = \lim_{x\to 0}\frac{\sin \pi x}{x} = \pi \lim_{x\to 0}\frac{\sin \pi x}{\pi x} = \pi

    Try similar things for the other ones.
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  6. #6
    Member sinewave85's Avatar
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    Ok, my somewhat indifferent math education may be catching up with me. I think I understand what you did, but I am still stuck on the next problems.

    <br /> <br />
L=\lim_{x \to 2} \frac {cos \frac{\pi} {x}} {x-2}<br /> <br />

    <br />
L=\lim_{t \to o} \frac {cos \frac{\pi} {t+2}} {t}<br />

    I get that I may need to use

    L=\lim_{t \to o} \frac {cos (t - 1)} {t} = 0

    but not how to get there. I have been working (unsuccessfully, obviously) at these all evening and may just be burned out, but I think I am missing something. Thanks for helping me in my hour of stupidity.
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sinewave85 View Post
    Ok, my somewhat indifferent math education may be catching up with me. I think I understand what you did, but I am still stuck on the next problems.

    <br /> <br />
L=\lim_{x \to 2} \frac {cos \frac{\pi} {x}} {x-2}<br /> <br />

    <br />
L=\lim_{t \to o} \frac {cos \frac{\pi} {t+2}} {t}<br />

    I get that I may need to use

    L=\lim_{t \to o} \frac {cos (t - 1)} {t} = 0

    but not how to get there. I have been working (unsuccessfully, obviously) at these all evening and may just be burned out, but I think I am missing something. Thanks for helping me in my hour of stupidity.
    Seeing that

    \cos\left(\frac{\pi}{2}\right)=0

    We can see we can rewrite this as

    \lim_{x\to{2}}\frac{\cos\left(\frac{\pi}{x}\right)-cos\left(\frac{\pi}{2}\right)}{x-2}

    So seeing that this fits the definition of the derivative at a point we have

    \lim_{x\to{2}}\frac{\cos\left(\frac{\pi}{x}\right)-\cos\left(\frac{\pi}{2}\right)}{x-2}=\bigg(\cos\left(\frac{\pi}{x}\right)\bigg)'\big  g|_{x=2}=\bigg(\frac{\pi\sin\left(\frac{\pi}{x}\ri  ght)}{x^2}\bigg)\bigg|_{x=2}=\frac{\pi}{4}
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sinewave85 View Post
    I am stuck on these problems, taking the limit at x = 2.

     f(x) = \frac {sin(\pi x)} {x-2}

     f(x) = \frac {cos \frac{\pi} {x}} {x-2}

     f(x) = \frac {\frac{1}{x} -1} {x-2}

    Any help is greatly appreciated, including general advice about working these kinds of problems. What do you do when you can't "see" how to rationalize an equation?
    \lim_{x\to{2}}\frac{\frac{1}{x}-1}{x-2}\cdot\frac{x}{x}=\lim_{x\to{2}}\frac{1-x}{x^2-2x}

    Now

    1-x=-(x-1)

    and

    x^2-2x=(x-1)^2-1

    So we have

    \lim_{x\to{2}}\frac{-(x-1)}{(x-1)^2-1}


    Now let x-1=t

    So as x\to{2}\Rightarrow{t\to{1}}

    so we have

    \lim_{t\to{1}}\frac{-t}{t^2-1}

    Which is obviosly undefined
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  9. #9
    Member sinewave85's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Seeing that

    \cos\left(\frac{\pi}{2}\right)=0

    We can see we can rewrite this as

    \lim_{x\to{2}}\frac{\cos\left(\frac{\pi}{x}\right)-cos\left(\frac{\pi}{2}\right)}{x-2}

    So seeing that this fits the definition of the derivative at a point we have

    \lim_{x\to{2}}\frac{\cos\left(\frac{\pi}{x}\right)-\cos\left(\frac{\pi}{2}\right)}{x-2}=\bigg(\cos\left(\frac{\pi}{x}\right)\bigg)'\big  g|_{x=2}=\bigg(\frac{\pi\sin\left(\frac{\pi}{x}\ri  ght)}{x^2}\bigg)\bigg|_{x=2}=\frac{\pi}{4}
    Ok, thanks for all of the help. But unfortunately, I am just getting more and more confused. I have not learned the derivative yet, which I realize must make this more difficult. Is there a way to get

    <br />
L=\lim_{x \to 2} \frac {cos \frac{\pi} {x}} {x-2}<br />

    into the form

    <br />
L=\lim_{x \to o} \frac {cos (x - 1)} {x} = 0<br />

    without using anything other than basic (precaluculus level) math? Like I said, I am at the very beginning of the course and don't know the definition of the derivative or l'H˘pital's rule. We're talking "calculus for dummies" level here -- apparently I have lost a few brain cells since the last math course I took .
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sinewave85 View Post
    Ok, thanks for all of the help. But unfortunately, I am just getting more and more confused. I have not learned the derivative yet, which I realize must make this more difficult. Is there a way to get

    <br />
L=\lim_{x \to 2} \frac {cos \frac{\pi} {x}} {x-2}<br />

    into the form

    <br />
L=\lim_{x \to o} \frac {cos (x - 1)} {x} = 0<br />

    without using anything other than basic (precaluculus level) math? Like I said, I am at the very beginning of the course and don't know the definition of the derivative or l'H˘pital's rule. We're talking "calculus for dummies" level here -- apparently I have lost a few brain cells since the last math course I took .
    FYI

    \lim_{x\to{0}}\frac{\cos(x-1)}{x}\ne{0}

    But

    \lim_{x\to{0}}\frac{1-\cos(x)}{x}=0\Rightarrow-\lim_{x\to{0}}\frac{\cos(x)-1}{x}=-0=0

    But hmm,

    I would suggest making a substitution that lets you expand the cosine by trig identites
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  11. #11
    Member sinewave85's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    \lim_{x\to{2}}\frac{\frac{1}{x}-1}{x-2}\cdot\frac{x}{x}=\lim_{x\to{2}}\frac{1-x}{x^2-2x}

    Now

    1-x=-(x-1)

    and

    x^2-2x=(x-1)^2-1

    So we have

    \lim_{x\to{2}}\frac{-(x-1)}{(x-1)^2-1}


    Now let x-1=t

    So as x\to{2}\Rightarrow{t\to{1}}

    so we have

    \lim_{t\to{1}}\frac{-t}{t^2-1}

    Which is obviosly undefined
    Thanks for the help on this one. I did most of the things you did, never got to that last step of putting it into a form that proves that it does not exist. I kept trying to make it exist, which apparently was the wrong tact!
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sinewave85 View Post
    Thanks for the help on this one. I did most of the things you did, never got to that last step of putting it into a form that proves that it does not exist. I kept trying to make it exist, which apparently was the wrong tact!
    You do know that for

    \lim_{x\to{c}}f(x) to exist we must have that

    \lim_{x\to{c^+}}f(x)=\lim_{x\to{c^{-}}}f(x)

    Right? Or in other words for a limit to exist we must have that the limit from the left equals the limit from the right

    So in this case

    \lim_{x\to{1^-}}\frac{-x}{x^2-1}=\infty

    whereas

    \lim_{x\to{1^+}}\frac{-x}{x^2-1}=-\infty

    \therefore\quad\lim_{x\to{1}}\frac{-x}{x^2-1}\text{ Does not exist}
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  13. #13
    Member sinewave85's Avatar
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    Ok, I have got it now. Thanks for the help, all.
    Last edited by sinewave85; June 28th 2008 at 12:22 PM.
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