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.

  • 1
    $\begingroup$ $|f(x)-g(x)| <\epsilon$ for $0<|x-a| <\delta $ does not imply that $f(x)=g(x)$. $\endgroup$ Commented Feb 24, 2021 at 5:24
  • 2
    $\begingroup$ $f(x)-g(x)$ 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. $\endgroup$ Commented Feb 24, 2021 at 5:24
  • $\begingroup$ 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 $x \to 0$. $\endgroup$ Commented Feb 24, 2021 at 7:24

3 Answers 3


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.)

  • $\begingroup$ 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 $\endgroup$
    – 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.


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