$f$ is a function with the domain of all real numbers, and let $a$, $L$ be elements of $\Bbb R$.
If $\lim_{x \rightarrow a}f(x) = L$, then $\lim_{x \rightarrow a/5}f(5x)=L$
How would you do this without the limit law theorems?
$f$ is a function with the domain of all real numbers, and let $a$, $L$ be elements of $\Bbb R$.
If $\lim_{x \rightarrow a}f(x) = L$, then $\lim_{x \rightarrow a/5}f(5x)=L$
How would you do this without the limit law theorems?
Take $\epsilon > 0$. We want $\delta > 0$ such that $$0< |x - a/5| <\delta \implies |f(5x)-L| < \epsilon,$$right? How do we find the little bastard? He's hidden inside the limit that we already know that exists. For that fixed $\epsilon$, there is $\delta' > 0$ satisfying $$0 < |t-a|<\delta'\implies |f(t) - L|<\epsilon.$$I used $t$ instead of $x$ on purpose. With this, put $\delta \doteq \delta'/5>0$. Then: $$0< |x-a/5| < \delta \implies 0 < |5x-a|<5\delta=\delta' \implies |f(5x)-L|<\epsilon,$$as wanted. The moral of the history is that $5x$ plays the role of $t$ on the limit we already know that holds. My other answer here might also be useful.