) = 0.8, page 70.(⇒) First assume that limx→∞f(x) =Land letxn∈(a,∞) such thatxn→ ∞asn→ ∞. We want to show thatf(xn)→L.Let>0. Since limx→∞f(x) =L, there existsM∈Rsuch that for anyx > M,|f(x)-L|<.Becausexn→ ∞, there exists a rankN0∈Nsuch that∀n≥N0,xn> M.Combining these, it follows that∀n≥N0,|f(xn)-L|<. Thusf(xn)→L.(⇐) Conversely, now assume thatf(xn)→Lfor any sequencexn∈(a,∞)such thatxn→ ∞, and we’d like to prove that limx→∞f(x) =L. We dothis using a proof by contradiction.Suppose thatf(x) does not convergetoLasxapproachesa.Then there exists an0>0 such that for anyM∈R, there exists a real numberxM(which depends onM) such thatxM> Mand|f(xM)-L| ≥0. In particular, for each natural numbern > a(taken asMin the above), we’ll find a real numberxn, such thatxn> nand|f(xn)-L| ≥0.The inequalityxn> nfor anyn > aimplies thatthe sequencexn→ ∞, thus by assumptionf(xn)→Lasn→ ∞.Thus|f(xn)-L|<0, fornsufficiently large. But this is in contradiction with|f(xn)-L| ≥0, for anyn∈N,n > a.
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