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f(x)=2x+1とすると，
∀ε>0, ∃δ>0 s.t. ∀x∈ℝ [|x-1|<δ⇒|f(x)-3|<ε].
（∵|f(x)-3|=|2x-2|=2|x-1|より，δ=ε/2をとればよい）
ゆえに，lim[x→1]f(x)=3.
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ありがとうございます！