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Math Help - limit problem with square root

  1. #1
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    limit problem with square root

    Hello, I am trying to solve the limit as x tends to 0 for this function:
    sqrt(|x|) * e^sin(pi/x)

    I am really not sure where to start here. Are you even allowed to have an absolute number in the square root. I thought that would be illegal unless you were using complex numbers.

    Thanks for any help with solving this.
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  2. #2
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    Re: limit problem with square root

    Quote Originally Posted by fran1942 View Post
    Hello, I am trying to solve the limit as x tends to 0 for this function:
    sqrt(|x|) * e^sin(pi/x)

    I am really not sure where to start here. Are you even allowed to have an absolute number in the square root. I thought that would be illegal unless you were using complex numbers.

    Thanks for any help with solving this.
    \displaystyle \begin{align*} \lim_{x \to 0}\sqrt{|x|}e^{\sin{\frac{\pi}{x}}} &= \lim_{x \to 0}\sqrt{|x|}\lim_{x \to 0}e^{\sin{\frac{\pi}{x}}} \\ &= \lim_{x \to 0}\sqrt{|x|}\lim_{x \to 0} \end{align*}

    Even though the second of these two limits doesn't exist, the first goes to 0, so no matter what value the second takes as you make x go closer to 0, it will end up multiplied by 0. Therefore the limit of the entire function is 0.
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: limit problem with square root

    Quote Originally Posted by Prove It View Post
    \displaystyle \begin{align*} \lim_{x \to 0}\sqrt{|x|}e^{\sin{\frac{\pi}{x}}} &= \lim_{x \to 0}\sqrt{|x|}\lim_{x \to 0}e^{\sin{\frac{\pi}{x}}} \\ &= \lim_{x \to 0}\sqrt{|x|}\lim_{x \to 0} \end{align*}

    Even though the second of these two limits doesn't exist, the first goes to 0, so no matter what value the second takes as you make x go closer to 0, it will end up multiplied by 0. Therefore the limit of the entire function is 0.
    Is it because the bound function  e^{\sin{\frac{\pi}{x}}}?
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  4. #4
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    Re: limit problem with square root

    Quote Originally Posted by Also sprach Zarathustra View Post
    Is it because the bound function  e^{\sin{\frac{\pi}{x}}}?
    YES,  \sqrt{|x|}~e^{-1}\le \sqrt{|x|}~e^{\sin{\frac{\pi}{x}}}\le \sqrt{|x|}~ e
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