Question-121725

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

$${f}\left({n}\right)=\frac{{n}^{\mathrm{2}} −\mathrm{2}{n}−\mathrm{15}}{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{5}{n}−\mathrm{3}}=\frac{{n}^{\mathrm{2}} −\mathrm{5}{n}+\mathrm{3}{n}−\mathrm{15}}{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{6}{n}−{n}−\mathrm{3}}=\frac{{n}\left({n}−\mathrm{5}\right)+\mathrm{3}\left({n}−\mathrm{5}\right)}{\mathrm{2}{n}\left({n}+\mathrm{3}\right)−\mathrm{1}\left({n}+\mathrm{3}\right)} \\ $$$${f}\left({n}\right)=\frac{\left({n}−\mathrm{5}\right)\left({n}+\mathrm{3}\right)}{\left({n}+\mathrm{3}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)}=\frac{{n}−\mathrm{5}}{\mathrm{2}{n}−\mathrm{1}} \\ $$$$\left.{i}\right){when}\:{n}\rightarrow\infty \\ $$$${li}\underset{{n}\rightarrow\infty} {{m}}\:\frac{{n}−\mathrm{5}}{\mathrm{2}{n}−\mathrm{1}} \\ $$$${li}\underset{{n}\rightarrow\infty} {{m}}\:\frac{\mathrm{1}−\frac{\mathrm{5}}{{n}}}{\mathrm{2}−\frac{\mathrm{1}}{{n}}}=\frac{\mathrm{1}−\mathrm{0}}{\mathrm{2}−\mathrm{0}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${when}\:{n}\rightarrow−\mathrm{3} \\ $$$${li}\underset{{n}\rightarrow−\mathrm{3}} {{m}}\:\frac{{n}−\mathrm{5}}{\mathrm{2}{n}−\mathrm{1}}=\frac{−\mathrm{3}−\mathrm{5}}{−\mathrm{6}−\mathrm{1}}=\frac{\mathrm{8}}{\mathrm{7}} \\ $$$${when}\:{n}\rightarrow\mathrm{0} \\ $$$${li}\underset{{n}\rightarrow\mathrm{0}} {{m}}\frac{{n}−\mathrm{5}}{\mathrm{2}{n}−\mathrm{1}}=\frac{−\mathrm{5}}{−\mathrm{1}}=\mathrm{5} \\ $$

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Question-121725

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