# What's wrong in this $\epsilon-\delta$ argument?

In Spivak's Calculus, one exercise asks whether the following is true:

Let $$f$$ and $$g$$ be functions such that $$f(x) < g(x)$$, for all $$x$$. Does it follow that $$\lim\limits_{x \to a} f(x) < \lim\limits_{x \to > a} g(x)$$?

The previous result holds if the signs $$<$$ get replaced by $$\leq$$, but it turns out this is not true in general for strict inequality. However I "proved" it was true. Obviously my argument is wrong, but it is not clear to me where lies the mistake, so I am requesting your help to figure it out.

First I envisioned using one neat trick I found in Terence Tao's blog (second paragraph in item 2), namely that to prove a quantity $$x$$ vanishes one can prove $$\lvert x \rvert \leq \epsilon$$, for every $$\epsilon > 0$$.

So my argument goes as follows: let $$f$$ and $$g$$ be functions as in the statement above. Then $$\lim\limits_{x \to a} f(x) \leq \lim\limits_{x \to a} g(x)$$. We show that equality leads to a contradiction.

If $$\lim\limits_{x \to a} f(x) = \lim\limits_{x \to a} g(x) = m$$ and $$\epsilon > 0$$, then there are $$\delta_1, \delta_2 > 0$$ such that

• if $$0 < \lvert x-a \rvert < \delta_1$$, we have $$\lvert f(x) - m \rvert < \cfrac{\epsilon}{2}$$;
• if $$0 < \lvert x-a \rvert < \delta_2$$, we have $$\lvert g(x) - m \rvert < \cfrac{\epsilon}{2}$$.

Now, if $$0 < \lvert x-a \rvert < \delta$$, where $$\delta$$ equals the smallest number between $$\delta_1$$ and $$\delta_2$$, then $$\lvert g(x) - f(x) \rvert = \lvert g(x) -m + m - f(x) \rvert \leq \lvert g(x) -m \rvert + \lvert m - f(x) \rvert < \cfrac{\epsilon}{2} + \cfrac{\epsilon}{2} = \epsilon$$.

This implies, by Prof. Tao's trick, that $$f(x) = g(x)$$; this is impossible since $$f(x) < g(x)$$ for all $$x$$, so we conclude $$\lim\limits_{x \to a} f(x) < \lim\limits_{x \to a} g(x)$$.

Where's the error? Thanks in advance.

• $|f(x)-g(x)| <\epsilon$ for $0<|x-a| <\delta$ does not imply that $f(x)=g(x)$. Commented Feb 24, 2021 at 5:24
• $f(x)-g(x)$ is not a fixed quantity. The issue with your argument is that $|f-g|<\epsilon$ has only been demonstrated for $x\in(a-\delta,a+\delta)$, a variable neighbourhood of $a$ due to $\delta$ being a function of $\epsilon$. Thus as $\epsilon$ becomes smaller we observe $\delta$ becoming smaller. Commented Feb 24, 2021 at 5:24
• Let me suggest an alternate approach. Take a simple counterexample and apply your "proof" to that counterexample. I suggest $f(x)= \vert x \vert$ for $x \neq 0, f(0)=1$, and $g(x)=0$ for all $x$, taking limits as $x \to 0$. Commented Feb 24, 2021 at 7:24

You can conclude that whenever $$x$$ satisfies $$0<|x-a|<\delta$$, you have $$|g(x)-f(x)|<\varepsilon$$. This is not enough to establish that $$f(x)=g(x)$$. In order to conclude that $$f(x)=g(x)$$, you'd need to know that for $$x$$ fixed, $$|f(x)-g(x)|<\varepsilon$$ for every $$\varepsilon>0$$. But this might not be true - there's no guarantee in general that for a smaller choice of $$\varepsilon$$ (say $$\varepsilon/2$$), the same $$\delta$$ is still small enough so that $$|x-a|<\delta\implies|f(x)-g(x)|<\varepsilon/2$$. You might need a smaller $$\delta$$, say $$\delta'$$, and it might not be the case that $$|x-a|<\delta'$$ anymore. (In fact, unless $$f$$ and $$g$$ are constant in a deleted neighborhood of $$a$$, this will never be the case.)

• Indeed. What @Iovita Kemény has shown is that, if we assume $f$ and $g$ have the same limit, then we can find $\delta$'s to make them arbitrarily close to each other, which is consistent with our assumption. To show that strict < doesn't necessarily hold for limits, Iovita might try to find a single example of $f(x)<g(x)$ everywhere but $\lim_{x\to a} f(x) = \lim_{x\to a} g(x)$. This may be easier than using a general argument. One such counterexample is given here: math.stackexchange.com/a/3968715/754927
– Ben
Commented Feb 24, 2021 at 5:43

Your $$\epsilon-\delta$$ argument is correct, but for the trick to work, your $$x$$ must be fixed. If a fixed non-negative quantity is $$\leq \epsilon$$ for any arbitrary $$\epsilon>0$$, then the quantity must be $$0$$. However, in this case $$|g(x)-f(x)|$$ is changing as $$x$$ takes any value in $$(0,\delta)$$.

You showed $$|g(x)-f(x)|<\epsilon$$ when $$x$$ is in the neighborhood $$0<|x-a|<\delta$$. This just means you can make $$g(x)$$ and $$f(x)$$ arbitrarily close when $$x$$ is not too far away from $$a$$, which is consistent with $$\displaystyle\lim_{x\to a} f(x)=\lim_{x\to a} g(x)$$.

That only shows it for that $$\epsilon$$. To show it for any other $$\epsilon$$ you might need another smaller $$\delta$$.

So for $$0 < |x-a| < \delta \implies |f(x) -g(x)| < \epsilon$$ for ALL $$\epsilon$$s you might need that $$0 < |x-a| < \delta$$ for ALL $$\delta$$. But that would mean $$0 < |x-a| = 0$$.

And it is true that for all $$x$$ where $$0 < |x-a| = 0$$ we will have $$f(x)= g(x)$$ but... there aren't any $$x$$ where $$0 < |x-a| = 0$$ so that is only vacuously true.