In formally proving that limx6 (x212x)=36\displaystyle \lim_{{{x}\to{6}}}\ {\left({x}^{{2}}-{12}{x}\right)}=-{36}, let ϵ>0\displaystyle \epsilon>{0} be arbitrary. Determine δ\displaystyle \delta as a function of ϵ\displaystyle \epsilon.

Note: in this case δ\displaystyle \delta will be a function of ϵ\displaystyle \epsilon. You will need to write the word epsilon for ϵ \displaystyle \ \epsilon\ in the answerbox


δ=\displaystyle \delta=