Contradiction on limit argument

Question

I'm sorry if this a stupid question. Could you point out the mistake in my reasoning?

The hypothesis says that $\lim_{x\to a} {f(x)} = {l}$ and $\lim_{x\to a} {g(x)} = {m}$

Which means that for every $\epsilon>0$ there is a $\delta>0$ such that if $0<\left|x-a\right|<\delta$, then $\left|f\left(x\right)-l\right|<\epsilon$ and $\left|g\left(x\right)-m\right|<\epsilon$

So

$l-\epsilon < f(x) < l+\epsilon$ and $m-\epsilon < g(x) < m+\epsilon $

But if you substract the inequalities you get

$(m-\epsilon)-(l-\epsilon) < g(x)-f(x) < (m+\epsilon)-(l+\epsilon) $

=

$m-l < f(x)-g(x) < m-l$

$m-l < m-l$

The only wrong step I could think of was eliminating $\epsilon$. If that is the case, could you explain why can't I do that?

Answer 1

You did not change the direction of the inequality as you subtract them. That is, it should be $$ (m-\epsilon)-(l+ \epsilon)<g(x)-f(x)<(m+ \epsilon)-(l-\epsilon)$$ instead.

Plus, it seems that a small step is omitted in your reasoning. (Don’t know if you intended to do so, but I decide to just point it out here)

By definition of limits, there exists $\delta_1$ such that $|x-a|<\delta_1$ implies $|f(x)-l|<\epsilon$ and there exists $\delta_2$ such that $|x-a|<\delta_2$ implies $|f(x)-m|<\epsilon$. Choose a positive $\delta$ smaller than both of them, then $|x-a|<\delta$ ensures both inequalities hold.