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Math Help - It's limit.

  1. #1
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    It's limit.

    Lim x->0 (e^sqrt(x) -2)/x
    solve it.
    Thank
    I solved it but I've got 2 answers and I'm not sure. Sorry about my english. I'm not a native speaker .
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  2. #2
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    Re: It's limit.

    Quote Originally Posted by Tiffany View Post
    Lim x->0 (e^sqrt(x) -2)/x solve it.
    I solved it but I've got 2 answers and I'm not sure.
    I hope the problem is actually {\lim _{x \to {0^ + }}}\frac{{{e^{\sqrt x }} - 2}}{x}, a one-sided limit.
    Does that limit exist? Hint: if it does, it is not bounded.
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  3. #3
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    Re: It's limit.

    Sorry my bad. It is -1 not -2
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  4. #4
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    Re: It's limit.

    Quote Originally Posted by Tiffany View Post
    Sorry my bad. It is -1 not -2
    That does not change much. It still must be a one-sided limit.
    The limit is unbounded.
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  5. #5
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    Re: It's limit.

    perhaps you were a bit over-eager to use l'hopital?

    we can use l'hopital once, to get:

    \lim_{x \to 0^+} \frac{e^{\sqrt{x}} - 1}{x} = \lim_{x \to 0^+} \frac{e^{\sqrt{x}}}{2\sqrt{x}}

    but in the latter limit, the numerator goes to 1, while the denominator goes to 0, so the ratio is unbounded.
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  6. #6
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    Re: It's limit.

    If you allow complex values, then it is alright to use a two-sided limit but then if you are allowing such things, you might as well consider the limit on the complex plane. This might be a bit of a tangent but here we go. Let

    f(z) = \frac{e^{\sqrt{z}}-1}{z}

    for complex values z. We'll show that f(z) is analytic. The condition with the Cauchy-Riemann equations is equivalent to saying that

    \frac{\partial f}{\partial \overline{z}} = \frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)f = 0

    for z = x + i y

    If you calculate this, it is indeed equal to zero. So the function is analytic and we can take the limit as z \to 0. The limit happens to be real and unbounded in the positive direction.
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