Let us verify using the definition of limit (with ε and δ ) that x→1lim(21∣x−1∣+3)=3. Let ε>0 be arbitrary. Put δ=ε. Then for each ε>0 there exists δ>0 such that if ∣x−1∣<δ then ∣(21∣x−1∣+3)−3∣=∣21∣x−1∣∣=21∣x−1∣<21δ=21ε<ε. Therefore,
x→1lim(21∣x−1∣+3)=3.
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