Limit of a Function Multiplied by a Scalar

In this post, I show that

$ \lim_{x \rightarrow a} [c f(x)] = c \lim_{x \rightarrow a} [f(x)]$

if $c$ is a variable for any real number. To do this, I approximately follow the steps in reference [1].

Essential Background Information:

Definition of a Limit:

A function $f(x)$ approaches a limit $A$ as $x$ approaches $a$ if, and only if, for each positive number $\epsilon$ there is another, $\delta$, such that whenever $0 < |x-a| < \delta $ we have $|f(x) – A| < \epsilon$. That is, when $x$ is near $a$ (within a distance $\delta$ from it), $f(x)$ is near $A$ (within a distance $\epsilon$ from it). In symbols we write $\lim_{x \rightarrow a} f(x) = A$.

Objective:

Suppose that $ |f(x) – A| < \epsilon$ whenever $0 < |x-a| < \delta $.

For $c \ne 0$, the objective is to show that

$ |cf(x) – cA| < \epsilon_1$ whenever $0 < |x-a| < \delta $,

in which $\epsilon_1$ is a positive number.

For $c=0$, the objective is to directly show that

$ \lim_{x \rightarrow a} [c f(x)] = c \lim_{x \rightarrow a} [f(x)]$.

Proof:

First consider the case of $c=0$. Suppose that each value of $f$ is a real number. Then

$ \lim_{x \rightarrow a} [0 f(x)] = \lim_{x \rightarrow a} [0] $

From this property,

$\lim_{x \rightarrow a} [0] = 0 $.

Assume that $A$ is a variable for any real number. Then

$0 = 0 \lim_{x \rightarrow a} [f(x)]$,

so it has been shown that for $c=0$

$\lim_{x \rightarrow a} [c f(x)] = c \lim_{x \rightarrow a} [f(x)]$.

Now suppose that $c$ is a nonzero real number.

Recall that $ |f(x) – A| < \epsilon$ whenever $0 < |x-a| < \delta $. Next multiply $ |f(x) – A| < \epsilon$ by $|c|$ to obtain $ |c| |f(x) – A| < |c| \epsilon$ whenever $0 < |x-a| < \delta $.

The next step is to use this absolute value property:

$ |c| |f(x) – A| = | c (f(x) – A) | = | c f(x) – cA |$.

Hence

$ | c f(x) – cA | < |c| \epsilon$ whenever $0 < |x-a| < \delta $.

Next define

$ |c| \epsilon \equiv \epsilon_1$. Note that $\epsilon_1 > 0$ as needed since $\epsilon > 0$ and $|c|>0$.

Then, if $c \ne 0$,

$ | c f(x) – cA | < \epsilon_1$ whenever $0 < |x-a| < \delta $,

with $\epsilon_1 >0$.

$\square$

References: