Prove the statement using the 𝜀, 𝛿 definition of a limit. lim x → 1 7 + 2x3 = 3Given 𝜀 > 0, we need 𝛿 such that if 0 < |x − 1| < 𝛿, then 7 + 2x3 − 3 . But 7 + 2x3 − 3 < 𝜀 ⇔ 2x − 23 < 𝜀 ⇔ 23|x − 1| < 𝜀 ⇔ |x − 1| < . So if we choose 𝛿 = , then 0 < |x − 1| < 𝛿 ⇒ 7 + 2x3 − 3 < 𝜀. Thus, lim x → 1 7 + 2x3 = 3 by the definition of a limit.
Question
Prove the statement using the 𝜀, 𝛿 definition of a limit. lim x → 1 7 + 2x3 = 3Given 𝜀 > 0, we need 𝛿 such that if 0 < |x − 1| < 𝛿, then 7 + 2x3 − 3 . But 7 + 2x3 − 3 < 𝜀 ⇔ 2x − 23 < 𝜀 ⇔ 23|x − 1| < 𝜀 ⇔ |x − 1| < . So if we choose 𝛿 = , then 0 < |x − 1| < 𝛿 ⇒ 7 + 2x3 − 3 < 𝜀. Thus, lim x → 1 7 + 2x3 = 3 by the definition of a limit.
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