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Thread: Limit problem

  1. #1
    Junior Member TriForce's Avatar
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    Limit problem

    Determine constant a so that lim f(x) when x->0 exists when

    f(x)={ arctan(1/x), x>0, (e^(ax)-1)/x, x<0 }


    The answer should be pi/2, but i don't understand the reasoning behind it.
    Anyone cares to explain?

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  2. #2
    MHF Contributor
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    Re: Limit problem

    Let's rewrite so it is easier to read:

    $\displaystyle f(x) = \begin{cases}\arctan\left( \dfrac{1}{x} \right) & x>0 \\ \dfrac{e^{ax}-1}{x} & x<0\end{cases}$

    Now, take the limit from the left and the right as $x \to 0$. You want them to be equal in order for the limit to exist. So:

    $\displaystyle \lim_{x \to 0^+}\arctan\left( \dfrac{1}{x} \right) = \dfrac{\pi}{2}$

    So, you want:

    $\displaystyle \dfrac{\pi}{2} = \lim_{x \to 0^-}\dfrac{e^{ax}-1}{x} = \lim_{x \to 0^-} ae^{ax} = a$

    It is actually extremely straightforward.
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  3. #3
    Junior Member TriForce's Avatar
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    Re: Limit problem

    Quote Originally Posted by SlipEternal View Post

    It is actually extremely straightforward.
    Thank you! That's extremely helpful, also thanks for formatting, i'll steal it in the future
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  4. #4
    Junior Member TriForce's Avatar
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    Re: Limit problem

    Sorry, my math is quite rusty, how did you lower the exponent a?

    $\displaystyle \dfrac{\pi}{2} = \lim_{x \to 0^-}\dfrac{e^{ax}-1}{x} = \lim_{x \to 0^-} ae^{ax} = a$
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  5. #5
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    Re: Limit problem

    Thanks from TriForce
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  6. #6
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    Re: Limit problem

    Or recognise the limit definition of the derivative at zero.
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