Limits of functions

**Pinkk** · April 26th 2010, 06:36 PM

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.

**Drexel28** · April 26th 2010, 07:41 PM

Originally Posted by Pinkk

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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