Find the limit of the sequence: a_n=\frac{8n^{2}+1n+1}{7n^{2}+9n+9}

osula9a 2022-01-04 Answered
Find the limit of the sequence:
\(\displaystyle{a}_{{n}}={\frac{{{8}{n}^{{{2}}}+{1}{n}+{1}}}{{{7}{n}^{{{2}}}+{9}{n}+{9}}}}\)

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Expert Answer

Raymond Foley
Answered 2022-01-05 Author has 1180 answers
We have
\(\displaystyle{a}_{{n}}={\frac{{{8}{n}^{{{2}}}+{1}{n}+{1}}}{{{7}{n}^{{{2}}}+{9}{n}+{9}}}}\)
Take \(\displaystyle{n}^{{{2}}}\) to be the common
\(\displaystyle{a}_{{n}}={\frac{{{n}^{{{2}}}{\left({8}+{\frac{{{n}}}{{{n}^{{{2}}}}}}+{\frac{{{1}}}{{{n}^{{{2}}}}}}\right)}}}{{{n}^{{{2}}}{\left({7}+{\frac{{{9}{n}}}{{{n}^{{{2}}}}}}+{\frac{{{9}}}{{{n}^{{{2}}}}}}\right)}}}}\)
\(\displaystyle{a}_{{n}}={\frac{{{8}+{\frac{{{1}}}{{{n}}}}+{\frac{{{1}}}{{{n}^{{{2}}}}}}}}{{{7}+{\frac{{{9}}}{{{n}}}}+{\frac{{{9}}}{{{n}^{{{2}}}}}}}}}\)
Apply limit
\(\displaystyle\lim_{{{n}\rightarrow\infty}}{a}_{{n}}=\lim_{{{n}\rightarrow\infty}}{\frac{{{8}+{\frac{{{1}}}{{{n}}}}+{\frac{{{1}}}{{{n}^{{{2}}}}}}}}{{{7}+{\frac{{{9}}}{{{n}}}}+{\frac{{{9}}}{{{n}^{{{2}}}}}}}}}\)
as \(\displaystyle{n}\rightarrow\infty\), \(\displaystyle{\frac{{{1}}}{{{n}}}}\rightarrow{0}\)
\(\displaystyle{n}\rightarrow\infty\), \(\displaystyle{\frac{{{1}}}{{{n}^{{{2}}}}}}\rightarrow{0}\)
\(\displaystyle{n}\rightarrow\infty\), \(\displaystyle{\frac{{{9}}}{{{n}}}}\rightarrow{0}\)
\(\displaystyle{n}\rightarrow\infty\), \(\displaystyle{\frac{{{9}}}{{{n}^{{{2}}}}}}\rightarrow{0}\)
\(\displaystyle\lim_{{{n}\rightarrow\infty}}{a}_{{n}}={\frac{{{8}+{0}+{0}}}{{{7}+{0}+{0}}}}\)
\(\displaystyle={\frac{{{8}}}{{{7}}}}\)
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