prove-the-limit-lim-x-2-2x-2-

Question Number 116819 by joki last updated on 07/Oct/20

p r o v e t h e l i m i t

l i \underset{x - ⟩ 2}{m} \sqrt{2 x} = 2

Answered by 1549442205PVT last updated on 07/Oct/20

Suppose 0 < ϵ < 2 be arbitrary small numer

so that ∣ \sqrt{2 x} - 2 ∣< ϵ \Leftrightarrow - ϵ + 2 < \sqrt{2 x} < ϵ + 2

\Leftrightarrow ϵ^{2} - 4 ϵ + 4 < 2 x < ϵ^{2} + 4 ϵ + 4

\Leftrightarrow \frac{ϵ^{2} - 4 ϵ + 4}{2} < x < \frac{ϵ^{2} + 4 ϵ + 4}{2}

\Leftrightarrow \frac{ϵ^{2} - 4 ϵ}{2} < x - 2 < \frac{ϵ^{2} + 4 ϵ}{2} . Then choosing

δ = \frac{4 ϵ - ϵ^{2}}{2} we need prove that \forall x so that

∣ x - 2 ∣< δ = \frac{4 ϵ - ϵ^{2}}{2} then ∣ \sqrt{2 x} - 2 ∣< ϵ . Ineed,

∣ x - 2 ∣< δ = \frac{4 ϵ - ϵ^{2}}{2} \Leftrightarrow - δ + 2 < x < 2 + δ

\Leftrightarrow - 2 δ + 4 < 2 x < 4 + 2 δ

\Rightarrow ϵ^{2} - 4 ϵ + 4 < 2 x < 4 ϵ - ϵ^{2} + 4 < 4 + 4 ϵ + ϵ^{2}

\Rightarrow 2 - ϵ < \sqrt{2 x} < ϵ + 2 \Rightarrow∣ \sqrt{2 x - 2} ∣< ϵ

That shows \lim_{x \to 2} \sqrt{2 x} = 2 (Q . E . D)