x↦5lim(5−x)21=∞x0=5,f(x)=(5−x)21.
It must be shown that
∀k>0∃δ(k)>0,∣x−x0∣<δ
performed
∣f(x)∣>k.
Consider
∣x−5∣<δ
then
∣x−5∣1>δ1
because
δ>0,∣x−5∣>0.
Consider
∣f(x)∣=∣∣(5−x)21∣∣=∣(x−5)2∣1==∣x−5∣21>δ21>k⟹δ21>k⟹δ2<k1⟹δ<k1.
So at δ<k1 have ∣∣(5−x)21∣∣>k ,
which means that
x↦5lim(5−x)21=∞.
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