##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> $\lim_{x\to a}\ f(x)=L$ if and only if $\lim_{x\to a^-}\ f(x)=L$ and $\lim_{x\to a^+}f(x)=L$.<br><b>Proof:</b> First we suppose that $\lim_{x\to a}\ f(x)=L$. This means for all $\epsilon>0$ there exists a $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon$. We want to show that $\lim_{x\to a^-}\ f(x)=L$ (the argument for $x\to a^+$ will be very similar). Consider any $\epsilon >0$. We must find a $\delta>0$ such that if $0<a-x<\delta$ then $|f(x)-L|<\epsilon$. By assumption we can find a $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon$. This $\delta$ will work for the desired conclusion since if $0<a-x<\delta$, then $0<|a-x|=|x-a|<\delta$, hence we get $|f(x)-L|<\epsilon$. This shows that $\lim_{x\to a^-}\ f(x)=L$, and as mentioned before, the argument for $x\to a^+$ is similar.<p> Now suppose that $\lim_{x\to a^-}\ f(x)=L$ and $\lim_{x\to a^+}\ f(x)=L$. This means for all $\epsilon>0$ there exists a $\delta_1>0$ such that if $0<a-x<\delta_1$ then $|f(x)-L|<\epsilon$, and there exists a $\delta_2$ such that if $0<x-a<\delta_2$ then $|f(x)-L|<\epsilon$. We want to show that $\lim_{x\to a}\ f(x)=L$. Consider any $\epsilon >0$. We must find a $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon$. Take $\delta=\text{min}(\delta_1,\delta_2)$ and assume $0<|x-a|<\delta$. Now either $x-a<0$ or $x-a\ge 0$. In the first case, $|x-a|=-(x-a)=a-x$, hence $0<a-x<\delta<\delta_1$, which implies $|f(x)-L|<\epsilon$. In the second case, $|x-a|=x-a$, hence $0<x-a<\delta<\delta_2$, which also implies $|f(x)-L|<\epsilon$. So in either case, we get $|f(x)-L|<\epsilon$, which shows that $\lim_{x\to a}\ f(x)=L$.