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Math Help - Limits of functions

  1. #1
    Senior Member Pinkk's Avatar
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    Limits of functions

    Let f be a function defined on some interval (0,a) and define g(y)=f(1/y) for y\in (a^{-1}, \infty). Show that \lim_{x\to 0^{+}}f(x) exists if and only if \lim_{y\to \infty}g(y) exists in which case they are equal.

    My attempt of proof:

    Let \lim_{x\to 0^{+}}f(x) = L and suppose L is finite. Then for each \epsilon > 0 there exists a > \delta > 0 such that x\in (0, \delta) implies |f(x)-L| <\epsilon. If x\in (0,\delta), then \frac{1}{x}\in (\delta^{-1},\infty), so \frac{1}{x}\in (\delta^{-1}, \infty) implies |f(x)-L|< \epsilon. Let y=\frac{1}{x}. So it follows that there exists \alpha < \infty, namely \alpha = \delta^{-1}, such that y > \alpha implies |f(\frac{1}{y}) - L | = |g(y) - L| <\epsilon, and so \lim_{y\to \infty} g(y) = \lim_{x\to 0^{+}}f(x)= L.

    The converse is pretty much the reverse of this proof and if L = +\infty , I'd just simply replace the epsilons with M > 0 and f(x) > M, and do the similar if L = -\infty

    I think this proof is correct, but I just want to make sure, thanks.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    Let f be a function defined on some interval (0,a) and define g(y)=f(1/y) for y\in (a^{-1}, \infty). Show that \lim_{x\to 0^{+}}f(x) exists if and only if \lim_{y\to \infty}g(y) exists in which case they are equal.

    My attempt of proof:

    Let \lim_{x\to 0^{+}}f(x) = L and suppose L is finite. Then for each \epsilon > 0 there exists a > \delta > 0 such that x\in (0, \delta) implies |f(x)-L| <\epsilon. If x\in (0,\delta), then \frac{1}{x}\in (\delta^{-1},\infty), so \frac{1}{x}\in (\delta^{-1}, \infty) implies |f(x)-L|< \epsilon. Let y=\frac{1}{x}. So it follows that there exists \alpha < \infty, namely \alpha = \delta^{-1}, such that y > \alpha implies |f(\frac{1}{y}) - L | = |g(y) - L| <\epsilon, and so \lim_{y\to \infty} g(y) = \lim_{x\to 0^{+}}f(x)= L.

    The converse is pretty much the reverse of this proof and if L = +\infty , I'd just simply replace the epsilons with M > 0 and f(x) > M, and do the similar if L = -\infty

    I think this proof is correct, but I just want to make sure, thanks.
    You have the correct idea, and the details I noticed seem correct. This is an easy proof, don't go looking for hidden complexity.
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