Is it true that $\lim \limits_{x \to a} [l(x)-m(x)] = 0 \implies l(x)$ has the same sign of $m(x)$ for $x$ sufficiently close to $a$? This was present in a demonstration of a theorem, however, it didn't seem obvious to me. Is this a general result?
Here is the exact context where this appeared (from Spivak's Calculus book):
If f is function for which $f(a) =f'(a) = f''(a) = ... = f^{n-1}(a) = 0$ and $f^{n}(a) > 0$, then $f$ will have a local minimum at $x=a$ if $n$ is even.
The demonstration uses a Taylor polynomial $$P_{n,a}(x) = \frac {f^{n}(a)}{n!}(x-a)^n$$
as $$\lim\limits_{x \to a} \frac{f(x)-P_{n,a}(x)}{(x-a)^n} =0$$
We would have 
$$\lim\limits_{x \to a} \left [ \frac{f(x)}{(x-a)^n} - \frac{f^{(n)}(a)}{n!} \right ] = 0$$
Which implies that $\frac{f(x)}{(x-a)^n}$ has the same sign as $\frac{f^{(n)}(a)}{n!}$, for $x$ sufficiently close to $a$.
As a consequence, $f(x)>0=f(a)$, because $f^{(n)}(a)>0$ and $(x-a)^n>0$, so there's a local minimum at $a$.
 A: First I will answer the question given as the title of your post.
If $\lim\limits_{x \to a}\{l(x) - m(x)\} = 0$ then it does not necessarily mean that $l(x)$ and $m(x)$ have the same sign as $x \to a$.
As as example let $ l(x) = x, m(x) = -x$ and $a = 0$ then clearly $l(x), m(x)$ have opposite signs as $x \to 0$.
However the context related to the analysis of maxima/minima via repeated derivatives is bit different from the title of your post. In the maxima/minima analysis you are given that $f^{(n)}(a) \neq 0$ and
$$\lim\limits_{x \to a} \left [ \frac{f(x)}{(x-a)^n} - \frac{f^{(n)}(a)}{n!} \right ] = 0$$ So in this case $l(x) = f(x)/(x - a)^{n}$ and $m(x) = f^{(n)}(a)/n!$ and the thing to note is that here $m(x)$ is a non-zero constant.
Thus we can rewrite the question as follows:

If $m$ is a non-zero constant and $\lim_{x \to a}\{l(x) - m\} = 0$ then does it mean that $l(x)$ is of the same sign as that of $m$ when $x \to a$?

Clearly the answer is Yes. Because this means that $\lim\limits_{x \to a}l(x) = m$ and thus as $x \to a$, $l(x)$ is very near $m$ and since $m \neq 0$ then $l(x)$ has same sign as that of $m$.
To see this it is important to forget the $\epsilon, \delta$ (because they tend to complicate simple stuff about inequalities and destroy so much the joy of learning calculus).
Let's assume that $m$ is positive and picture $m$ on number line as follows:
-------------------------$0$---------------$m$-----------------------
As $x \to a$ the values of $l(x)$ will come close to $m$ and in the definition of limit we are guaranteed that values of $l(x)$ will come as close to $m$ as we want and hence for values of $x$ sufficiently near $a$ we should ultimately have one of the following pictures
-------------------------$0$-------$l(x)$--------$m$-----------------------
-------------------------$0$---------------$m$-------$l(x)$----------------
In both cases we can see that $l(x)$ is positive. A more concrete argument is like this: suppose that $l(x)$ remains negative (or zero) so that it is to the left of $0$ on number line. In that case the gap between $l(x)$ and $m$ will always be greater than (or equal to) $m$ and hence values of $l(x)$ can't come as close to $m$ as we want. This contradiction shows that values of $l(x)$ must be ultimately positive as $x \to a$.
If $m < 0$ we can modify the above figures for number line in an obvious manner and all the action takes place to the left of $0$ on number line.
A: Here is how I would say things with a few more details: since $\lim\limits_{x\to a}\dfrac{f(x)}{(x-a)^n}=\dfrac{f^{(n)}(a)}{n!}$, we have, for any $\varepsilon >0$, 
$$\dfrac{f^{(n)}(a)}{n!}-\varepsilon\leq \dfrac{f(x)}{(x-a)^n}\leq\dfrac{f^{(n)}(a)}{n!}+\varepsilon $$
as soon as $x$ is close enough to $a$.
Now choose $\varepsilon$ so that $\dfrac{f^{(n)}(a)}{n!}-\varepsilon\geq 0$. Then you have $\dfrac{f(x)}{(x-a)^n}\geq 0$, which implies $f(x)\geq 0$ since $n$ is even.
A: It does need signal and only use limit and continuity. Generally, $l(x)-M(x) \to0$ does not mean $l(x)$ and $m(x)$ must have same derivatives of various orders, and it is easy to find many examples. The problem can be proved as follow. 
First by Taylor series
$$f(x)=f(a)+\dfrac{f^{(n)}(t)} {n!}(x-a)^n$$
where $|t-a|<|x-a|$. Since for all $x$ that $|x-a|<\delta$, 
$f^{(n)}(t)>0$ and  $(x-a)^n>0$. So we have
$$
\lim \limits_{x \to a}\dfrac{f^{(n)}(t)} {n!}(x-a)^n \geqslant 0 
$$
Note: $\lim \limits_{x \to a}f^{(n)}(t)=f^{(n)}(a)$. 
So for all $x$ that $|x-a|<\delta$, $f(x)-f(a)\geqslant0$ or $f(x)\geqslant f(a)$. 
Thus $a$ is local minimum. 
