Proof $\lim_{x \rightarrow \infty}f(x)=\lim_{x \rightarrow 0^{+}}f(\frac{1}{x})$

Solution 1:

I have not checked each detail, but the idea seems correct. The underlying principle is:

$\lim_{x\to 0,x>0}g(x) = L$ iff for each $\epsilon>0$ there is a $\delta$ so that $g((0,\delta))\subset B_\epsilon(L)$. $\lim_{x\to\infty} h(x) = L$ iff for each $\epsilon>0$ we have a $\rho$ so that $h((\rho,\infty))\subset B_\epsilon(L)$.

Now the basic idea is that the transformation $x\to 1/x$ maps each $(0,\delta)$ bijectively to some $(\rho,\infty)$ with $\rho=1/\delta$.

Note that an even simpler proof can be done using this characterisation of limit: We say $\lim_{x\to x_0} f(x) = L$ iff for any sequence $x_n$ with $x_n\to x_0$ we have $f(x)\to L$ ($x_0$ can be $\pm\infty$). It is clear that this is equivalent to the other idea of convergence: Trivially if $\lim_{x\to x_0}f(x)=L$ by the first characterisation then also $\lim_{n\to \infty} f(x_n)=L$ for any $x_n\to x_0$. On the other hand if the first characterisation is not met one can easily construct a sequence $x_n\to x_0$ so that $|f(x_n)-L|\geq \epsilon$ for some $\epsilon>0$.

So if we use this characterisation your property follows directly from $$ x_n \to 0 \Leftrightarrow 1/x_n \to \infty$$ for $x_n>0$.