In formally proving that limx6 (x2+x)=42\displaystyle \lim_{{{x}\to{6}}}\ {\left({x}^{{2}}+{x}\right)}={42}, let ϵ>0\displaystyle \epsilon>{0} be arbitrary. Choose δ = \displaystyle \delta\ =\ min(ϵm , 1)\displaystyle {\left(\frac{\epsilon}{{m}}\ ,\ {1}\right)}. Determine the smallest value of m\displaystyle {m} that would satisfy the proof.

m= \displaystyle {m}=\