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Math Help - Prove this limit

  1. #1
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    Prove this limit

    We just learned this (well, a much more basic version) today in calculus, and I have no idea where to start on this problem.

    <br />
\lim_{x \to 0^{+}} \sqrt{x}*e^{\sin(\pi/x)} = 0<br /> <br />

    "
    your task is to first find two outer functions that "squeeze" the given one to the right of zero; then compute the appropriate limits and make your conclusion"



    EDIT:

    http://www.mathhelpforum.com/math-he...e-theorem.html

    i searched and found the above thread (which made me laugh because it's the same problem!), but im still very lost

    thanks
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by silencecloak View Post
    We just learned this (well, a much more basic version) today in calculus, and I have no idea where to start on this problem.

    <br />
\lim_{x \to 0^{+}} \sqrt{x}*e^{\sin(\pi/x)} = 0<br /> <br />

    "
    your task is to first find two outer functions that "squeeze" the given one to the right of zero; then compute the appropriate limits and make your conclusion"



    EDIT:

    http://www.mathhelpforum.com/math-he...e-theorem.html

    i searched and found the above thread (which made me laugh because it's the same problem!), but im still very lost

    thanks
    \forall t \in \mathbb{R} ~,~ -1 \le \sin(t) \le 1
    What if you let t=\frac \pi x ?

    Then compose with the exponential function (which is increasing) and then multiply by \sqrt{x} (which is positive).
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    \forall t \in \mathbb{R} ~,~ -1 \le \sin(t) \le 1
    What if you let t=\frac \pi x ?

    Then compose with the exponential function (which is increasing) and then multiply by \sqrt{x} (which is positive).
    Not gonna lie I have no idea what the first part of that math equation says

    I do understand that  -1 \le \sin(t) \le 1

    and that this is true even while
    t=\frac \pi x

    but that is as far as I can understand
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  4. #4
    Super Member 11rdc11's Avatar
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    The first part just means sin(t) has the domain of all real numbers

    We also know

    -1 \le \sin(t) \le 1

    So make it look like your original function by putting the number e in

    e^{-1} \le e^{sin(t)} \le e^{1}

    Can you do the next step?
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  5. #5
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    Quote Originally Posted by 11rdc11 View Post
    The first part just means sin(t) has the domain of all real numbers

    We also know

    -1 \le \sin(t) \le 1

    So make it look like your original function by putting the number e in

    e^{-1} \le e^{sin(t)} \le e^{1}
    and then substitute "t" with \pi/x ?

    and that's it?

    but then i only want the lim as x tends to 0 from the right, so i would leave off the whole -1 side

    thats not to bad, just have to wrap my head around it
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  6. #6
    Moo
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    Quote Originally Posted by silencecloak View Post
    and then substitute "t" with \pi/x ?

    and that's it?

    thats not to bad, just have to wrap my head around it
    Yes, since it is okay for any value of t
    I used t so that you could see more easily, but I think it rather confused you
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  7. #7
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    Because the sine function is bounded by -1 below and 1 above we have:
    e^{ - 1}  \le e^{\sin \left( {\frac{\pi }{x}} \right)}  \le e.

    So we have this \sqrt x e^{ - 1}  \le \sqrt x e^{\sin \left( {\frac{\pi }{x}} \right)}  \le \sqrt x e.

    But \lim _{x \to 0^ +  } \sqrt x  = 0 so the limit is 0.
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  8. #8
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    ok so for the grand finally, my final answer is

    "your task is to first find two outer functions that "squeeze" the given one to the right of zero; then compute the appropriate limits and make your conclusion"

    the two outer functions are

    <br /> <br />
\sqrt x e^{ - 1}  \le \sqrt x e^{\sin \left( {\frac{\pi }{x}} \right)}  \le \sqrt x e^{1}<br />


    and the limits of each function, lets call them h(x), f(x), g(x) all = 0
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  9. #9
    Super Member 11rdc11's Avatar
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    Yep
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  10. #10
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    so are the two squeezing functions always going to resemble the function they are squeezing?

    in this case, the middle function had \sqrt{x} and "e" in it, so we made the squeezing functions to look like this also
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  11. #11
    Super Member 11rdc11's Avatar
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    Not necessarily. I can't think of an example off the top of my head but I know when working with sequences and using the squeeze theorem it doesn't necessarily have to resemble the function. Let me see if I come up with an example.
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  12. #12
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    Quote Originally Posted by 11rdc11 View Post
    Not necessarily. I can't think of an example off the top of my head but I know when working with sequences and using the squeeze theorem it doesn't necessarily have to resemble the function. Let me see if I come up with an example.
    Might it be easier to just explain why in this case the two other functions did resemble the original in this way?
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  13. #13
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    Quote Originally Posted by silencecloak View Post
    so are the two squeezing functions always going to resemble the function they are squeezing?
    That would be nice but not necessary.
    Consider: \lim _{x \to 0} \left| x \right|\left( {\arctan \left( {\frac{\pi }{x}} \right)} \right).
    Because  - \frac{\pi }{2} \le \arctan (x) \le \frac{\pi }{2} we have  - \frac{\pi }{2}\left| x \right| \le \left| x \right|\arctan \left( {\frac{\pi }{x}} \right) \le \left| x \right|\frac{\pi }{2}.

    Now the limit is clear.
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  14. #14
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    Quote Originally Posted by Plato View Post
    That would be nice but not necessary.
    Consider: \lim _{x \to 0} \left| x \right|\left( {\arctan \left( {\frac{\pi }{x}} \right)} \right).
    Because  - \frac{\pi }{2} \le \arctan (x) \le \frac{\pi }{2} we have  - \frac{\pi }{2}\left| x \right| \le \left| x \right|\arctan \left( {\frac{\pi }{x}} \right) \le \left| x \right|\frac{\pi }{2}.

    Now the limit is clear.
    Ok, so at least the way I'm seeing it, the two squeezing functions still do in a way relate to the original. In the way that sin relates to -1 and 1 arctan relates to -\pi/2 and \pi/2

    And that's pretty much as far as the resemblance between the original and squeezing functions go
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  15. #15
    Super Member 11rdc11's Avatar
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    Here is an example

    \lim_{x \to \infty} \frac{1}{e^{x}}

    You could use squeeze thoerem for this.

    f(x) = 0

    g(x) = \frac{1}{e^x}

    h(x) = \frac{1}{x}

    Prob not the best example but you can see what I mean lol
    Last edited by 11rdc11; September 15th 2008 at 03:30 PM. Reason: oops to late
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