How do I formally prove this by using epsilon and delta?

$a\neq 0$, if $\lim \limits_{x \to 0}f(x)= L$, then $\lim \limits_{x \to \infty}f(\frac{a}{x})= L$

  • $\begingroup$ Hint: you can make a substitution $t = \frac{a}{x}$ $\endgroup$ – Kevin Sheng Dec 16 '14 at 23:39
  • $\begingroup$ must be there a relation between delta and $|f(\frac{a}{x})-L|< \epsilon$ ? $\endgroup$ – Firas Ali Abdel Ghani Dec 17 '14 at 0:23

Suppose $ \lim \limits_{x \to 0}f(x)= L$. Let $\epsilon \gt 0$ be an arbitrary positive quantity.

Our aim here is to find a real number $M$ such that if $x \gt M$ then $ |f(\frac a x) - L | \lt \epsilon $. But we can use the fact that $\frac a x $ can be made sufficiently large if we make $x$ sufficiently small or in other words sufficiently close to $0$.

Hence, since $ \lim \limits_{x \to 0}f(x)= L$, there exists $\delta \gt 0$ such that $ x \in (- \delta , \delta) \implies |f(x) - L| \lt \epsilon $. So all we need to do is to let $M = \frac{|a|}{\delta} \in \Bbb R$. Then,

$$ x \gt \frac{|a|}{\delta} \gt 0 \implies \delta \gt \frac{|a|}{x} = |\frac a x - 0| \implies \frac a x \in (- \delta , \delta) \implies |f(\frac a x) - L | \lt \epsilon $$

$ \mathscr {Q.E.D.}$

  • $\begingroup$ Brilliant! I didn't think about the using the relation between $x>M>0$ !!thank you! $\endgroup$ – Firas Ali Abdel Ghani Dec 17 '14 at 7:37
  • $\begingroup$ but when we say that $ |\frac a x - 0| \implies \frac a x \in (- \delta , \delta$ it doesn't mean that we are talking about specific area and not in convergence in infinity? $\endgroup$ – Firas Ali Abdel Ghani Dec 17 '14 at 7:39
  • $\begingroup$ @FirasAliAbdelGhani: I'm sorry don't understand your question mate? The bounding of $x$ beyond $M$, that is sending $x$ to infinity, is what results in the confinement of $\frac a x $, to the $\delta$-"area". $\endgroup$ – Ishfaaq Dec 17 '14 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.