# Scalar product rule for limits

Let $X$ be a metric space, $E$ be a subset of a metric space $X$, $p$ be a limit point of $E$, $f:E \to \mathbb{C}$, and $\lim\limits_{x \to p} f(x)=L$. Finally let $c \in \mathbb{C}$. Show that $\lim\limits_{x \to p} cf(x)=cL$.

Let $\epsilon > 0$ be given. Want to show that there exists $\delta>0$ such that $|cf(x)-cL|<\epsilon$ whenever $|x-p|<\delta$.

Suppose $c\neq 0$.
Then $|cf(x)-cL|< \epsilon \Rightarrow |c||f(x)-L|<\epsilon \Rightarrow |f(x)-L|<\frac{\epsilon}{|c|}$.

I'm stuck on showing that $\delta>0$ exists. Or, I may be potentially doing this proof wrong.

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## 1 Answer

From $\lim\limits_{x \to p} f(x)=L$, given $\epsilon > 0$, there exists $\delta >0$ such that $0< |x-p|<\delta$ implies $|f(x)-L|<\epsilon$.

So for given $\varepsilon >0$, let $\epsilon =\frac{\varepsilon}{|c|}$ where $c\neq 0$. Then $0 < |x-p|<\delta$ implies $|cf(x)-cL|<|c|\epsilon =\varepsilon$.

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