Can we always replace $\underset{\delta\rightarrow 0}{\lim}$ with $\underset{k\delta\rightarrow 0}{\lim}$ for $k$ constant?

Question

Is it in general acceptable to replace $\underset{\delta\rightarrow 0}{\lim}$ with $\underset{k\delta\rightarrow 0}{\lim}$ for constant $k$?

What I mean by this is, if we have a real-valued function $f$ and we know that, say $\underset{\delta \rightarrow 0}{\lim} f(\delta)=L$ for real constant $k$, is it then valid to say that $$\underset{\delta \rightarrow 0}{\lim} f(k \delta)=\underset{k\delta \rightarrow 0}{\lim} f(k \delta)=\underset{\delta \rightarrow 0}{\lim} f(\delta)=L\:?$$

(I feel like this should always be the case, because for any constant $k$, $(\delta\rightarrow 0) \Longleftrightarrow (k\delta \rightarrow 0)$ but maybe I'm overlooking something)

Answer 1

You can do this from scratch: Change $\delta$ to $x$ for clarity. Fix $k\neq 0.$ Suppose $\underset{x \rightarrow 0}{\lim} f(x)=L$. Then, let $\epsilon>0.$ There is a $\delta>0$ so that if $|x|<\delta$ then $|f(x)-L|<\epsilon.$ Suppose such a $\delta$ has been produced and that $|kx|<|k|\delta.$ Set $\delta'=|k|\delta$. Then, if $|kx|<\delta'$ we have $|kx|<|k|\delta\Rightarrow |x|<\delta\Rightarrow |f(x)-L|<\epsilon$. So, with $\delta'$ answering the $\epsilon$-challenge, we have by definition, $\underset{kx \rightarrow 0}{\lim} f(x)=L$.

Answer 2

Yes you can!

$\forall k \in \Bbb{R}\backslash \{0\}$,

$$\underset{\delta \rightarrow 0}{\lim} f(\delta)=\underset{k\delta \rightarrow 0}{\lim} f(k \delta)=\underset{t \rightarrow 0}{\lim} f(t)=L$$

where $t=k\delta$

Sometimes be careful with the sign of $k$. It may change a RHL to LHL or vice versa and this creates a problem when the function is not continous