[tex]\bf \lim\limits_{x\to 4}~2x-1=7~\hfill \begin{cases} 0<|x-4|<\delta\\[-0.5em] \hrulefill\\ |(2x-1)-7|<\epsilon \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ |(2x-1)-7|<\epsilon\implies |2x-8|<\epsilon\implies 2|x-4|<\epsilon\implies |x-4|<\stackrel{\downarrow }{\cfrac{\epsilon}{2}} \\\\[-0.35em] ~\dotfill\\\\ 0|x-4|<\delta\implies 0<|x-4|<\stackrel{\downarrow }{\cfrac{\epsilon}{2}} \\\\[-0.35em] ~\dotfill\\\\ 2|x-4|<2\delta\implies 2|x-4|<2\left( \cfrac{\epsilon}{2} \right)\implies 2|x-4|<\epsilon[/tex]
just to clarify some, is a cheesy proof, since it's circular, but the epsilon-delta proofs are circular and thus most often cheesy.
