I want to prove that given $f$ and $g$ such that for all $x$, $f(x) \leqslant g(x)$, then the limit of $f$ at $a$ is less or equal than the limit of $g$ at $a$, provided they both exist. Let us call them respectively $l_1$ and $l_2$
So far, I have done this. We have a $\delta$ for any $\varepsilon$ such that: if $|x-a| < \delta$, then both $|f(x)-l_1| < \varepsilon$ and $|g(x) -l_2| < \varepsilon$ hold. Now, by the reverse triangle inequality I have said that this means that $|f(x)| - |l_1| < \varepsilon$ and $|g(x)| - |l_2| < \varepsilon$. Now, I don’t know what to do to compare l1 and l2 effectively. Please, don’t spoil the answer, just give me some useful and justified hints.