Prove if $\lim_{x\to 0}$ $f(x)=L$ then $\lim_{x\to 0} f(ax)=L$ for non-zero constant $a$

Question

Does it require two different cases for $a>0$ and $a<0$? Am I allowed to use $f(ax) = af(x)$ here? Thanks. Keeping in mind I am a beginner and this is my first post. Appreciate the help

Answer 1

Use the $\varepsilon-\delta$ language. $\lim_{x \to 0} f(X)=L$, says that for every $\varepsilon$ there is a $\delta$ such that $|x-0|=|x|< \delta \implies |f(x)-L|<\varepsilon$. Now, examine $|ax|<\delta$ and go from there.

Solution

Let $\lim_{x \to 0} f(X)=L$. Then for every $\varepsilon$ there is a $\delta$ such that $|x-0|=|x|< \delta \implies |f(x)-L|<\varepsilon$.

Now, for $\varepsilon$ choose the $\delta$ given by the above assumption and let $|ax| < \delta$. Now, let $y=ax$ and so $|y|<\delta$. Then, $|f(y)-L|<\varepsilon$. So, if we consider our function to be $f(ax)$, then choosing $\delta'=\frac{\delta}{|a|}$ we have $|x|<\delta' \implies |f(ax)-L|< \varepsilon $

In the case where $a=0$, then $ax=0$ for all $x$ and so $f(0)$ is either defined or it isn't and the result would follow immediately depending on which case.

Answer 2

You are given $\displaystyle\lim_{X \to 0} f(X)=L$. By the definition of limits, this means that given $\epsilon>0$, there is a $\delta>0$ such that if $|X-0|=|X|<\delta$, then $|f(X)-L|<\epsilon$.

Remember this definition is just saying that given an error, $\epsilon$, you can find a number, $\delta$, such that if $x$ is no 'farther' than $\delta$ from $0$, i.e. $-\delta<x<\delta$, then $f(x)$ is no 'farther' from the number $L$ than $\epsilon$, i.e. $L-\epsilon<f(x)<L+\epsilon$.

Your problem is to do the same for $f(ax)$, where $a$ is a fixed nonzero number. So given error $\epsilon>0$, you need to find a number, let's say $D>0$, so that if $|x-0|=|x|<D$, then $|f(ax)-L|<\epsilon$.

Given a value $X$, let $x=X/a$, i.e. $X=ax$. (Notice this is allowed since $a \neq 0$). We know from above that given $\epsilon>0$, we can find $\delta>0$ such that if $|X-0|=|X|<\delta$ then $|f(X)-L|<\epsilon$. But then $|X|<\delta$ means that $|ax|<\delta$, i.e. $|a||x|<\delta$, so that $|x|<\delta/|a|$. Is almost what we wanted! Let $D=\delta/|a|$. Then we know that if $|x|<D=\delta/|a|$, then $|f(X)-L|<\epsilon$. But $|f(ax)-L|=|f(X)-L|<\epsilon$. Then we have found the $D$ we wanted! Success!