 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-ii/2/a/iv/solution

Given

ε > 0

, the two assumed limits of sequences supply thresholds

N_{1}, N_{2}

such that

∣ a_{n} - ℓ ∣ < ε

for

n \geq N_{1}

and

∣ c_{n} - ℓ ∣ < ε

for

n \geq N_{2}

. For

n \geq max {N_{1}, N_{2}}

ℓ - ε < a_{n} \leq b_{n} \leq c_{n} < ℓ + ε .

Thus

∣ b_{n} - ℓ ∣ < ε

eventually, proving the squeeze theorem:

b_{n} ⟶ ℓ .

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