# Find error in proof for $f(x) < g(x) \implies \lim_{x\to a}f(x) < \lim_{x\to a}g(x)$

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 $x\neq 0,\quad g(0) = 1$
$a=0$

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 $L_1 < L_2$.

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

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 $x \rightarrow 0$.
• 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 $\delta$? May 4, 2016 at 22:59