Let ε>0 be fixed. We may assume that ε⩽21 , otherwise we take δ for ε=21 . Solve
∣∣21x2−x+1−1∣∣<ε⇔∣∣x2−2x∣∣<2ε⇔⇔{x2−2x−2ε<0x2−2δδx+2ε>0⇔{x∈(1−1+2ε,1+1+2ε)x∈(−∞,1−1−2ε)∪(1+1−2ε,+∞)
Thus
x∈(1+1−2ε,1+1+2ε)⇒∣∣21x2−x+1−1∣∣<ε
Let δ=min(1+2ε−1,1−1−2ε)>0
Then x∈(2−δ,2+δ)⇒∣∣21x2−x+1−1∣∣<ε
which by the definition proves
x→2lim(21x2−x+1)=1
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