Lim x → 4 1 + 1 4 x = 2 given ε > 0, we need δ such that if 0 < |x − 4| < δ, then 1 + 1 4 x − 2 . but 1 + 1 4 x − 2 < ε ⇔ 1 4 x − 1 < ε ⇔ 1 4 |x − 4| < ε ⇔ |x − 4| < . so if we choose δ = then 0 < |x − 4| < δ ⇒ 1 + 1 4 x − 2 < ε. thus, lim x → 4 1 + 1 4 x = 2 by the definition of a limit.