Question

using ϵ−δ definition, show that

lim _x→2 (1/2 x² - x + 1)

Answer 1

Let $\varepsilon >0$ be fixed. We may assume that $\varepsilon \leqslant \frac{1}{2}$ , otherwise we take $\delta$ for $\varepsilon =\frac{1}{2}$ . Solve

$\left| \frac{1}{2}x^2-x+1-1 \right|<\varepsilon \Leftrightarrow \left| x^2-2x \right|<2\varepsilon \Leftrightarrow \\\Leftrightarrow \left\{ \begin{array}{c} x^2-2x-2\varepsilon <0\\ x^2-2\delta δ x+2\varepsilon >0\\\end{array} \right. \Leftrightarrow \left\{ \begin{array}{c} x\in \left( 1-\sqrt{1+2\varepsilon},1+\sqrt{1+2\varepsilon} \right)\\ x\in \left( -\infty ,1-\sqrt{1-2\varepsilon} \right) \cup \left( 1+\sqrt{1-2\varepsilon},+\infty \right)\\\end{array} \right.$

Thus

$x\in \left( 1+\sqrt{1-2\varepsilon},1+\sqrt{1+2\varepsilon} \right) \Rightarrow \left| \frac{1}{2}x^2-x+1-1 \right|<\varepsilon$

Let $\delta =\min \left( \sqrt{1+2\varepsilon}-1,1-\sqrt{1-2\varepsilon} \right) >0$

Then $x\in \left( 2-\delta ,2+\delta \right) \Rightarrow \left| \frac{1}{2}x^2-x+1-1 \right|<\varepsilon$

which by the definition proves

$\underset{x\rightarrow 2}{\lim}\left( \frac{1}{2}x^2-x+1 \right) =1$