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Thread: Limits of functions

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

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

    My attempt of proof:

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

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

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

    My attempt of proof:

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

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

    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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