Archives for Pure Math - Page 3
Given conditions: lim β‘ π‘ β π π ( π‘ ) = π lim tβa β g(t)=b lim β‘ π₯ β π π ( π₯ ) = π lim xβb β f(x)=c To prove: lim β‘ π‘ β π π ( π ( π‘ ) ) = π lim tβa β f(g(t))=c
Given conditions: limβ‘tβag(t)=b\lim_{t \to a} g(t) = blimtβaβg(t)=b limβ‘xβbf(x)=c\lim_{x \to b} f(x) = climxβbβf(x)=c To prove: limβ‘tβaf(g(t))=c\lim_{t \to a} f(g(t)) = climtβaβf(g(t))=c Using the definition of the limit: limβ‘tβag(t)=b\lim_{t \to…