##The lines of the proof have been shuffled, and your task is to sort them into the logically correct order (if there seem to be several correct orders, choose one that produces the 'most sensible' proof). Click a round button <input type=radio> to select a line, then click the appropriate square grey button <input type=button value=" > "> to move the line to that place. When you think you are done, click on the Grade button.<p> ##<b>Theorem:</b> Let $a<b$ be real numbers, and suppose $f(x)\le g(x)$ for all $x\in(a,b)$. If $\lim_{x\to a}\ f(x)=L$ and $\lim_{x\to a}\ g(x)=M$ then $L\le M$.<br><b>Proof:</b> Suppose the assumptions are satisfied, but $L>M$. We have to show that this is impossible, which means we need to derive a contradiction. By the Limit Law for differences, $\lim_{x\to a}\ (g(x)-f(x))=M-L$. Since $L-M>0$ by assumption, we can take this to be $\epsilon$ and deduce that there exists a $\delta>0$ such that if $0<|x-a|<\delta$ then $|g(x)-f(x)-(M-L)|<L-M$. Since $r\le |r|$ for any real number $r$, we deduce that if $0<|x-a|<\delta$ then $g(x)-f(x)-M+L<L-M$, hence $g(x)< f(x)$. Let $c=\text{min}\{b,a+\delta\}$, and choose any $x'\in(a,c)$. Then $0<x'-a=|x'-a|<\delta$, so it follows that $g(x')< f(x')$. But this contradicts the assumption that $f(x)\le g(x)$ for all $x\in(a,b)$. Therefore $L>M$ is impossible, and the theorem is proved.