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Math Help - find limit

  1. #1
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    find this limit

    limit ( e^(-1/x^2)/x,x,tends to infinity)
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  2. #2
    GAMMA Mathematics
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    Quote Originally Posted by szpengchao View Post
    limit ( e^(-1/x^2)/x,x,tends to infinity)
    It goes to zero. Why?

    x^2 goes to infinity

    \frac{1}{x^2} then goes to zero

    e^{\frac{-1}{x^2}} approaches one

    \frac{1}{x} approaches zero
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  3. #3
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    Maybe he meant \lim_{x\to 0 }\frac{\exp (-1/x^2)}{x}.
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  4. #4
    Math Engineering Student
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    Quote Originally Posted by szpengchao View Post
    limit ( e^(-1/x^2)/x,x,tends to infinity)
    By setting u=\frac1{x^2} your limit is 0.
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  5. #5
    Eater of Worlds
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    Here's a method. I hope it will suffice. Kind of one sided though.

    \lim_{x\rightarrow{0}}\frac{e^{\frac{-1}{x^{2}}}}{x}

    Let t=e^{\frac{-1}{x^{2}}}

    Then you get:

    \lim_{t\rightarrow{0^{+}}}t\sqrt{-ln(t)}=0

    You can always try the "ol' Hospital" rule.
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  6. #6
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    L'H˘pital's rule seems to be a dead end for this problem. The second form ends up as \mathop {\lim }\limits_{x \to 0} \frac{{( - 1)( - 2)x^{ - 3} e^{\frac{{ - 1}}{{x^2 }}} }}{1} \implies \mathop {\lim }\limits_{x \to 0} \frac{{2e^{\frac{{ - 1}}{{x^2 }}} }}{{x^3 }}. I don't know if there is any analytical way to solve this without the use of a substitution.
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