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Math Help - How should i prove this?

  1. #1
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    How should i prove this?

    I'm stuck after i tried to integrate by parts.

    \lim  x  \int_x^1 \frac{cos t}{t^2} dt = 1

    The limit is actually as x \rightarrow 0^+ , I can't seemed to key in the input.

    Thanks in advance.

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  2. #2
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    Quote Originally Posted by xcluded View Post
    I'm stuck after i tried to integrate by parts.

    \lim  x  \int_x^1 \frac{cos t}{t^2} dt = 1

    The limit is actually as x \rightarrow 0^+ , I can't seemed to key in the input.

    Thanks in advance.

    I wonder if the comparison test would work, which you find a function similar to
    the function which in this case you can do, someone correct me if I am wrong
    since cosine is bounded by -1 and 1
      -1\le cos(t)\le1
    so then
      \frac{-1}{t^2}\le \frac{cos(t)}{t^2}\le \frac{1}{t^2}
    so in comparsion, you can say you function behaves like the function
    \frac{1}{t^2}, if you dont know the comparsion test, look it up it can prove helpful, if you find a function larger than your original then you the function you were given which is small will also converge, if you find a function smaller than the original diverges then the original will diverge, this does not apply to vice versa so now you can do

    \lim_{x\to0+}\int_x^1\frac{1}{t^2}

    \lim_{x\to0+}-\frac{1}{t}\|_x^1

    you should be able to take it from here
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by xcluded View Post
    I'm stuck after i tried to integrate by parts.

    \lim x \int_x^1 \frac{cos t}{t^2} dt = 1

    The limit is actually as x \rightarrow 0^+ , I can't seemed to key in the input.

    Thanks in advance.

    Is this \lim_{x\to0^+}x\int_x^1\frac{\cos(z)}{z^2}dz or \lim_{x\to0^+}\int_x^1\frac{\cos(z)}{z^2}dz

    Seeing your asnwer I assume it's the first.

    I hate using it, but I assume it's how this is to be done

    \lim_{x\to0^+}\frac{\int_x^1\frac{\cos(z)}{z^2}dz}  {\frac{1}{x}}

    Both the denominator and numerator approach infinity as x\to0^+. Apply L'hopital's rule and...done.
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  4. #4
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    Quote Originally Posted by RockHard View Post
    I wonder if the comparison test would work, which you find a function similar to
    the function which in this case you can do, someone correct me if I am wrong
    since cosine is bounded by -1 and 1
      -1\le cos(t)\le1
    so then
      \frac{-1}{t^2}\le \frac{cos(t)}{t^2}\le \frac{1}{t^2}
    so in comparsion, you can say you function behaves like the function
    \frac{1}{t^2}, if you dont know the comparsion test, look it up it can prove helpful, if you find a function larger than your original then you the function you were given which is small will also converge, if you find a function smaller than the original diverges then the original will diverge, this does not apply to vice versa so now you can do

    \lim_{x\to0+}\int_x^1\frac{1}{t^2}

    \lim_{x\to0+}-\frac{1}{t}\|_x^1

    you should be able to take it from here
    Hmm what about the x outside the integral ?
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    Is this \lim_{x\to0^+}x\int_x^1\frac{\cos(z)}{z^2}dz or \lim_{x\to0^+}\int_x^1\frac{\cos(z)}{z^2}dz

    Seeing your asnwer I assume it's the first.

    I hate using it, but I assume it's how this is to be done

    \lim_{x\to0^+}\frac{\int_x^1\frac{\cos(z)}{z^2}dz}  {\frac{1}{x}}

    Both the denominator and numerator approach infinity as x\to0^+. Apply L'hopital's rule and...done.
    i got this

     \lim_{x\to0^+}\frac{x^2 sinx - 2x cos x}{x^2}

    but never got it equal to 1.

    i did something wrong?
    Last edited by xcluded; November 18th 2009 at 08:30 PM.
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  6. #6
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by xcluded View Post
    i got this

     \lim_{x\to0^+}\frac{x^2 sinx - 2x cos x}{x^2}

    but never got it equal to 1.

    i did something wrong?
    \frac{d}{dx}\left[\int_x^1\frac{\cos t}{t^2}\,dt\right]=-\frac{\cos x}{x^2}

    \frac{d}{dx}\left[\frac{1}{x}\right]=-\frac{1}{x^2}

    So...
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  7. #7
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    Quote Originally Posted by redsoxfan325 View Post
    \frac{d}{dx}\left[\int_x^1\frac{\cos t}{t^2}\,dt\right]=-\frac{\cos x}{x^2}

    \frac{d}{dx}\left[\frac{1}{x}\right]=-\frac{1}{x^2}

    So...
    OH ! Ok i did something terribly wrong earlier.
    Thanks !
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  8. #8
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    Was my idea somewhat? I am hoping to make some sort of progress here, lol. So If I ever need correct, feel free too even if your the OP
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  9. #9
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    Quote Originally Posted by RockHard View Post
    Was my idea somewhat? I am hoping to make some sort of progress here, lol. So If I ever need correct, feel free too even if your the OP
    Yea my lecturer used the comparison test to do it similar question too.
    By the way , do i need to compare once only ? As in ,

    \lim_{x\to0+}\int_x^1\frac{1}{t^2} = 1<br />

    Hence the whole thing converges to 1 ?
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  10. #10
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by RockHard View Post
    Was my idea somewhat? I am hoping to make some sort of progress here, lol. So If I ever need correct, feel free too even if your the OP
    You would need to squeeze it on both sides, not just one.

    \lim_{x\to0^+}x\int_x^1-\frac{1}{t^2}\,dt\leq\lim_{x\to0^+}x\int_x^1\frac{  \cos t}{t^2}\,dt\leq\lim_{x\to0^+}x\int_x^1\frac{1}{t^2  }\,dt

    So \lim_{x\to0^+}x\cdot\frac{x-1}{x}\leq\lim_{x\to0^+}x\int_x^1\frac{\cos t}{t^2}\,dt\leq\lim_{x\to0^+}x\cdot\frac{1-x}{x}

    So all this tells you is that -1\leq\lim_{x\to0^+}x\int_x^1\frac{\cos t}{t^2}\,dt\leq1

    Or maybe since it goes to zero from the positive side, 0\leq\lim_{x\to0^+}x\int_x^1\frac{\cos t}{t^2}\,dt\leq1
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