Proof via epsilon delta.

Question

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

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

Answer 1

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.}$