##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 $b<a<c$ be real numbers, suppose $f(x)\le g(x)\le h(x)$ for all $x\in(b,a)\cup(a,c)$ and $\lim_{x\to a}\ f(x)=L=\lim_{x\to a}\ h(x)$. Then $\lim_{x\to a}\ g(x)=L$.<br><b>Proof:</b> Suppose the assumptions above are satisfied, and consider any $\epsilon >0$. To show that $\lim_{x\to a}\ g(x)=L$, we must find a $\delta>0$ such that if $0<|x-a|<\delta$ then $|g(x)-L|<\epsilon$. Since $\lim_{x\to a}\ f(x)=L$ we can find a $\delta_1>0$ such that if $0<|x-a|<\delta_1$ then $|f(x)-L|<\epsilon$. Since $\lim_{x\to a}\ g(x)=M$ we can find a $\delta_2>0$ such that if $0<|x-a|<\delta_2$ then $|g(x)-M|<\epsilon$. Note that the conclusions in the previous two sentences are equivalent to $L-\epsilon<f(x)<L+\epsilon$ and $L-\epsilon<h(x)<L+\epsilon$. Take $\delta=\text{min}\{\delta_1,\delta_2\}$ and assume $0<|x-a|<\delta$. Then $0<|x-a|<\delta_1$ and $0<|x-a|<\delta_2$, so $L-\epsilon<f(x)\le g(x)\le h(x)<L+\epsilon$. Hence we have shown that if $0<|x-a|<\delta$ then $|g(x)-L|<\epsilon$. This proves that $\lim_{x\to a}\ g(x)=L$.