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Question Number 121725 by oustmuchiya@gmail.com last updated on 11/Nov/20

Answered by TANMAY PANACEA last updated on 11/Nov/20

f(n)=((n^2 −2n−15)/(2n^2 +5n−3))=((n^2 −5n+3n−15)/(2n^2 +6n−n−3))=((n(n−5)+3(n−5))/(2n(n+3)−1(n+3))) f(n)=(((n−5)(n+3))/((n+3)(2n−1)))=((n−5)/(2n−1)) i)when n→∞ lim_(n→∞) ((n−5)/(2n−1)) lim_(n→∞) ((1−(5/n))/(2−(1/n)))=((1−0)/(2−0))=(1/2) when n→−3 lim_(n→−3) ((n−5)/(2n−1))=((−3−5)/(−6−1))=(8/7) when n→0 lim_(n→0) ((n−5)/(2n−1))=((−5)/(−1))=5

f(n)=n2−2n−152n2+5n−3=n2−5n+3n−152n2+6n−n−3=n(n−5)+3(n−5)2n(n+3)−1(n+3)f(n)=(n−5)(n+3)(n+3)(2n−1)=n−52n−1i)whenn→∞limn→∞n−52n−1limn→∞1−5n2−1n=1−02−0=12whenn→−3limn→−3n−52n−1=−3−5−6−1=87whenn→0limn→0n−52n−1=−5−1=5

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