I know that it is not true that $f(x) < g(x) \implies \lim_{x\to a}f(x) < \lim_{x\to a}g(x)$

A counter example could be
$f(x) = 0$
$g(x) = |x|$ if $x\neq 0,\quad g(0) = 1$

However, before I thought about it I came up with the following proof and now I can't see what is wrong with it:

Let $\lim_{x\to a}f(x) = L_1, \quad \lim_{x\to a}g(x) = L_2$
We use proof by contradiction, so assume $L_1 \geq L_2$

As a first case, assume $L_1 = L_2 = L$.
Since $f(x) < g(x)$ we can set $d_g = g(x) - f(x) > 0$
We know there is a $\delta$ such that, if we set $\epsilon = \dfrac{d_g}{2}$

$|x-a| < \delta \implies \left\{ \begin{array}{ll} |f(x) - L| < \dfrac{d_g}{2} \implies L - \dfrac{d_g}{2} < f(x) < L + \dfrac{d_g}{2} \\ |f(x) + d_g - L| < \dfrac{d_g}{2} \implies L - \dfrac{3d_g}{2} < f(x) < L - \dfrac{d_g}{2} \end{array} \right.$

So $f(x) < L - \frac{d_g}{2} < f(x)$. A contradiction, so $L_1 \neq L_2$

As a second case, assume $L_1 > L_2$.
Set $d_L = L_1 - L_2 > 0$
We know there is a $\delta$ such that, if we set $\epsilon = \dfrac{d_L}{2}$

$|x-a| < \delta \implies \left\{ \begin{array}{ll} |f(x) - (L_2 + d_L) | < \dfrac{d_L}{2} \implies L + \dfrac{d_L}{2} < f(x) < L + \dfrac{3d_L}{2} \\ |g(x) - L_2| < \dfrac{d_L}{2} \implies L - \dfrac{d_L}{2} < g(x) < L+ \dfrac{d_L}{2} \end{array} \right.$

So $g(x) < L + \dfrac{d_L}{2} < f(x)$. A contradiction, so $L_1 \leq L_2$

The two cases together shows that $L_1 < L_2$.

  • 2
    $\begingroup$ $d_g$ can't be independent of $x$ in the example, because in fact they get close together near $x=0$. $\endgroup$
    – Ian
    May 4, 2016 at 22:52

1 Answer 1


You cannot correlate a constant $d_g$ to be independent of the functions you equate them with. They have a straight relationship and they collapse together when $x \rightarrow 0$.

  • $\begingroup$ Oh, I see. So it is because in the definition of a limit, strictly speaking, $x$ is introduced after $\delta$ and thus can't be used to define $\delta$? $\endgroup$ May 4, 2016 at 22:59
  • $\begingroup$ Yes, exactly. When you are using a definition and generally in mathematics, you are strictly defining units. So, d as a constant being independent is a mistake and of course cannot be used. $\endgroup$
    – Rebellos
    May 4, 2016 at 23:01
  • $\begingroup$ By the way, remember please to approve the questions. An answered question should be go as answered and approved thread if it fits you (you forgot to do that in the previous limit question I fully answered you as well). $\endgroup$
    – Rebellos
    May 4, 2016 at 23:04
  • $\begingroup$ I know that, but you can not approve at once and in the previous question I did not understand your explanation so I just upvoted and left the question for a while to see if I can understand your explanation later. $\endgroup$ May 4, 2016 at 23:07
  • $\begingroup$ I edited it. The proof in your book had typo error between a and 0. Go check it please :) $\endgroup$
    – Rebellos
    May 4, 2016 at 23:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .