Difficulties with Spivak limits problem.Help appreciated Basically this is the problem and I can not find sense in the solution written in accompanying solution manual.
Problem goes like this:


Suppose that $ f(x) \le g(x) $ for all x.Prove that $$ \lim_{x \to a}f(x) \le \lim_{x \to a}g(x)$$ 


Solution given in manual goes like this:


Assume that $$ l=\lim_{x \to a}f(x) \gt \lim_{x \to a}g(x) = m$$ $$\epsilon=l-m\gt0$$  $$\delta\gt0$$
if $0\lt|x-a|\lt\delta $ then $|l-f(x)|\lt \frac\epsilon2 $ and $|m-g(x)|\lt \frac\epsilon2$
Thus for $0\lt|x-a|\lt\delta$ we have $$\epsilon\lt m +\frac\epsilon2 = l- \frac\epsilon2 \lt f(x)$$


What confuses me the most if how does he obtain that $$\epsilon\lt m +\frac\epsilon2 = l- \frac\epsilon2 \lt f(x)$$ which is a contradiction
Reference : Michael Spivak Calculus chapter 5 exercise 12.a
 A: Something seems to be wrong with your solution manual. $\epsilon < m+\epsilon/2 \iff m > \epsilon/2 = (l-m)/2 \iff 3m > l$, so that the contradiction argument would be invalid when $3m \leq l$. That last line should probably read $$g(x) < m+\frac{\epsilon}{2} = l -\frac{\epsilon}{2} < (\mbox{you fill here}),$$ which leads to a contradiction.
A: Assume that both $f$ and $g$ are defined in a (punctured) neighborhood $\dot U$ of $a$ and that $\lim_{x\to a} f(x)=\alpha$, $\>\lim_{x\to a} g(x)=\beta$. 
Given any $\epsilon>0$ there is a $\delta>0$ such that $0<|x-a|<\delta$ implies both of
$$f(x)>\alpha-\epsilon, \qquad
 g(x)<\beta+\epsilon\ .$$
Fix any such $x$. Then
$$\beta+\epsilon>g(x)\geq f(x)>\alpha-\epsilon\ ,$$
or $\beta-\alpha>-2\epsilon$. Since this is true for any $\epsilon>0$ we necessarily have $\beta-\alpha\geq0$.
A: As pointed out, the claim $\epsilon<m+\dfrac{\epsilon}{2}$ doesn't make sense since this requires $m$ to be greater than $\dfrac12{(l-m)}$, thus putting restrictions on $m$.
To arrive at $g(x)<m+\dfrac{\epsilon}2=l-\dfrac{\epsilon}2<f(x)$, observe:
\begin{align}
&g(x)-m\leq|g(x)-m|<\dfrac{\epsilon}2 \\\\&\quad\quad\quad\text{and similarly}, \\\\&l-f(x)\leq|l-f(x)|<\dfrac{\epsilon}2.
\end{align}
A: I have checked the manual and Spivak wrote "$g(x)< m +\varepsilon /2 = l+\varepsilon/2 < f(x) $" you have a typo.
Maybe this fact  is much more easily to see as follows: Suppose that $a$ is and adherent point on $S$ which is a subset of the domain of $f,g$. Suppose that $\lim_{x \to a; x\in S}f(x)=L$ and $\lim_{x \to a; x\in S} g(x)= M$ and $g(x)\le f(x)$ for all the $x$ in $S$. Then $M\le L$. 
Let define $h(x)=f(x)-g(x)$ so $h(x)\ge 0$ for all the $x\in S$ and by the limit laws we can conclude that $\lim_{x \to a; x\in S}h(x)=L-M$. Let $N=L-M$. If we can show that $N\ge 0$ we're done. Suppose to the contrary that $N<0$ and let $\varepsilon = |N|/2$. Then there is a $\delta >0$ such that for $\{x\in S:|x-a|< \delta\}$ we have $|h(x)-N|<|N|/2$. Thus, $h(x)< |N|/2+N=N/2$, so $h(x)<0$, a contradiction.
Note that the above argument works for $g(x)<f(x)$ (why?)
