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Math Help - Limit x->0+ X^(2/x) HELP!

  1. #1
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    Limit x->0+ X^(2/x) HELP!

    I need know how to show the work on this limit problem. I have been to several school tutors who can't seem to do it correctly. What is the limit as x approaches 0 from the right of x^(2/x)? The question as you would type it into an Computer Algebra System is: limit(x^(2/x),x,0,1)
    This gives us 0 which is the correct answer but I have no idea how to show the work. I know that in the end it someone should be e^(-∞) which equals 0. Someone please help!
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  2. #2
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    Re: Limit x->0+ X^(2/x) HELP!

    Have you considered the logarithm? log\left(x^{\frac{2}{x}}\right) = \frac{2}{x}\cdot log(x)
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  3. #3
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    Re: Limit x->0+ X^(2/x) HELP!

    Yeah I got there. It makes sense now. I guess it ends up being -∞ because 2ln(.0000000000...1) is a large negative number over a very very small number. Thank you for the reminder. I made the mistake of trying to use L'Hospitals rule since that was the chapter this questions was in :/.
    Thanks!
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  4. #4
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    Re: Limit x->0+ X^(2/x) HELP!

    Do you know L'H˘pital's rule?

    Write \displaystyle x^{2/x}\text{   as   }\left(e^{\ln(x)\right) ^{2/x}=\left(e\right)^{\displaystyle\frac{2\ln(x)}{x}}

    The exponential function is continuous, so \lim_{x\to a}(e^{f(x)})=\left(e\right)^{ \lim_{x\to a}(f(x))}

    Use L'H˘pital's rule to evaluate the limit, \lim_{x\to0^+}\frac{2\ln(x)}{x}}
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  5. #5
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    Re: Limit x->0+ X^(2/x) HELP!

    I didn't think I could use L'Hopital's rule because I can't seem to get the correct indeterminate form (∞/∞, -∞/-∞, ect..).
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  6. #6
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    Re: Limit x->0+ X^(2/x) HELP!

    Quote Originally Posted by NeedDirection View Post
    I didn't think I could use L'Hopital's rule because I can't seem to get the correct indeterminate form (∞/∞, -∞/-∞, ect..).
    Doh!

    Yup, that's \frac{=\infty}{0^+}, which is -∞ , so that's good!
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