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](https://terrytao.wordpress.com/2010/10/21/245a-problem-solving-strategies/) 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.