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

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

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  • $\begingroup$ 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
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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.}$

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  • $\begingroup$ 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

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