$\displaystyle \lim_{x\rightarrow 1} \frac{2+4x}{3} = 2 $

$\displaystyle 0<\left | x-1 \right | < \delta $

$\displaystyle \left | \frac{2+4x}{3}-2 \right | <\epsilon $

simplified i get $\displaystyle \left | x-4 \right |< \frac{\epsilon }{4} $

i set $\displaystyle \frac{\epsilon }{4} = \delta $

then i get $\displaystyle 0< \left | x-1 \right | < \frac{\epsilon }{4} $

I dont have an answer key for that problem in my calculus book so i was wondering if i did it right?

Thankss