Answer to Question #313536 in Calculus for betty

Question #313536

using ϵ−δ definition, show that

lim x→2 (1/2 x2 - x + 1)


1
Expert's answer
2022-03-19T02:39:41-0400

Let ε>0\varepsilon >0 be fixed. We may assume that ε12\varepsilon \leqslant \frac{1}{2} , otherwise we take δ\delta for ε=12\varepsilon =\frac{1}{2} . Solve

12x2x+11<εx22x<2ε{x22x2ε<0x22δδx+2ε>0{x(11+2ε,1+1+2ε)x(,112ε)(1+12ε,+)\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(1+12ε,1+1+2ε)12x2x+11<εx\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 δ=min(1+2ε1,112ε)>0\delta =\min \left( \sqrt{1+2\varepsilon}-1,1-\sqrt{1-2\varepsilon} \right) >0

Then x(2δ,2+δ)12x2x+11<εx\in \left( 2-\delta ,2+\delta \right) \Rightarrow \left| \frac{1}{2}x^2-x+1-1 \right|<\varepsilon

which by the definition proves

limx2(12x2x+1)=1\underset{x\rightarrow 2}{\lim}\left( \frac{1}{2}x^2-x+1 \right) =1


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