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Math Help - Limits & Indeterminate Forms

  1. #1
    Newbie superphysics's Avatar
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    Exclamation Limits & Indeterminate Forms

    Hello. I need help on two questions related to limits of indeterminate forms. I've solved at least fifty other such questions, but these just elude me...

    1. \lim_{x\rightarrow\infty}(\sinh(x)-x)
    2. \lim_{x\rightarrow0}\frac{x-\tan^{-1}(x)}{x\sin(x)}

    Much obliged if someone can help!
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  2. #2
    Eater of Worlds
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    \lim_{x\rightarrow{\infty}}[sinh(x)-x]

    \lim_{x\rightarrow{\infty}}[\frac{e^{x}}{2}-\frac{1}{2e^{x}}-x]={\infty}
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by galactus View Post
    \lim_{x\rightarrow{\infty}}[sinh(x)-x]

    \lim_{x\rightarrow{\infty}}[\frac{e^{x}}{2}-\frac{1}{2e^{x}}-x]={\infty}
    ooops, not true since you have the case infinity - infinity..
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  4. #4
    Super Member PaulRS's Avatar
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    For the second:

    Remember that:

    \arctan(x)=\sum_{i=0}^{n}{\frac{(-1)^i\cdot{x^{2i+1}}}{2i+1}}+E_n

    Where: |E_n|<\frac{x^{2n+1}}{2n+1}

    and : \lim_{x\rightarrow{0}}{\frac{\sin(x)}{x}}=1

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  5. #5
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by superphysics View Post
    Hello. I need help on two questions related to limits of indeterminate forms. I've solved at least fifty other such questions, but these just elude me...

    1. \lim_{x\rightarrow\infty}(\sinh(x)-x)
    2. \lim_{x\rightarrow0}\frac{x-\tan^{-1}(x)}{x\sin(x)}

    Much obliged if someone can help!
    for number 2, use L'hospital's Rule twice since it is the case of 0/0. you should get 0.
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  6. #6
    Super Member PaulRS's Avatar
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    Quote Originally Posted by kalagota View Post
    ooops, not true since you have the case infinity - infinity..

    e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...

    Then: e^x-x=1+\frac{x^2}{2}+\frac{x^3}{3!}+...

    Taking the limit: \lim_{x\rightarrow{<br />
+\infty}}{e^x-x}=+\infty

    Another way: \lim_{x\rightarrow{<br />
+\infty}}{e^x-x}=\lim_{x\rightarrow{<br />
+\infty}}{x\cdot{( \frac{e^x}{x}-1} )} and you can check by L' H˘pital that: \lim_{x\rightarrow{<br />
  +\infty}}{\frac{e^x}{x}}=+\infty
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  7. #7
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by PaulRS View Post
    e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...

    Then: e^x-x=1+\frac{x^2}{2}+\frac{x^3}{3!}+...

    Taking the limit: \lim_{x\rightarrow{<br />
+\infty}}{e^x-x}=+\infty

    Another way: \lim_{x\rightarrow{<br />
+\infty}}{e^x-x}=\lim_{x\rightarrow{<br />
+\infty}}{x\cdot{( \frac{e^x}{x}-1} )} and you can check by L' H˘pital that: \lim_{x\rightarrow{<br />
+\infty}}{\frac{e^x}{x}}=+\infty
    i know that it is the answer, yet in his solution, i think, any prof would still argue that some arguments are missing.. ▄ anyways, our goal is to help ryt? ▄
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  8. #8
    Eater of Worlds
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    I'm sorry I wasn't more thorough, but I had to leave in a hurry in the middle, so I gave an abridged version.
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  9. #9
    Newbie superphysics's Avatar
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    I've run into another problem. This one, unlike the other two, which were purely for interest, is for my course study.

     <br />
\lim_{x\rightarrow\infty}\frac{1-\cos{x}}{x}<br />

    My problem here is that I can't figure out a way to handle \cos{x}, which oscillates between -1 & 1 when x approaches \infty. And we are not allowed to use any series expansions, which may help solve the problem.

    Please?
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  10. #10
    Super Member PaulRS's Avatar
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    To deal with that sort of limits you can use the Squeeze Theorem

    See here: Squeeze theorem - Wikipedia, the free encyclopedia
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  11. #11
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    Quote Originally Posted by superphysics View Post
    I've run into another problem. This one, unlike the other two, which were purely for interest, is for my course study.

     <br />
\lim_{x\rightarrow\infty}\frac{1-\cos{x}}{x}<br />

    My problem here is that I can't figure out a way to handle \cos{x}, which oscillates between -1 & 1 when x approaches \infty. And we are not allowed to use any series expansions, which may help solve the problem.

    Please?
    Multiply the numerator and denominator by 1+\cos x and use the limit \sin x / x \to 0.
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