# Limit of difference of two functions

I've come across different proofs for one of the limit laws, namely, $$lim_{x\to c} [f - g](x) = lim_{x \rightarrow c}\ f(x) - lim_{x \rightarrow c}\ g(x) .$$

Now, some authors end their proof like so,

\begin{aligned}|[f-g](x)-(L-M)| & =|(f(x)-L)-(g(x)-M)| \\ & \leq|f(x)-L|+\mid g(x)-M \\ & <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.\end{aligned}

while others like this,

\begin{aligned}|[f-g](x)-(L-M)| & =|(f(x)-L)-(g(x)-M)| \\ & \leq|f(x)-L|+\mid g(x)-M \\ & <\epsilon+\epsilon=2\epsilon.\end{aligned} Can someone please explain the reasoning behind, $$\epsilon/2$$ or just $$\epsilon$$ ?

• It is just a matter of taste. Feb 25 at 11:03
• As long as you can make the estimate as small as you wish it is fine. So $2\varepsilon,$ $\sqrt{\varepsilon}$ and many other expressions are admissible. But not $\varepsilon^\varepsilon.$ It may look weird but also $\varepsilon^{-1}$ is OK. Feb 25 at 19:02

When you prove the existence of a limit using the $$\epsilon$$-$$\delta$$ definition, you are proving a statement of the form "for every $$\epsilon>0$$, there exists a $$\delta >$$ such that..." This means that the first line of your proof should be something like "Let $$\epsilon >0$$ be arbitrary..." Since $$\epsilon$$ is chosen arbitrarily, the quantity $$2\epsilon$$ is also arbitrary.
If you like, you can think of the two epsilons in your two arguments as being different, say $$\epsilon_1, \epsilon_2$$, and notice that the following holds: for every $$\epsilon_1>0$$, there exists $$\epsilon_2>0$$ such that $$2\epsilon_2 = \epsilon_1$$.