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 $\epsilon$ ?

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    $\begingroup$ It is just a matter of taste. $\endgroup$ Feb 25 at 11:03
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    $\begingroup$ 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. $\endgroup$ Feb 25 at 19:02

1 Answer 1


It's not clear to me if this is the "reasoning" you are hoping is explained, but here goes.

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 $2\epsilon_2 = \epsilon_1$.

Which should you choose? Like Peter said in the comment, it's a matter of taste.


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