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

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

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Expert Answer

lenkiklisg7
Answered 2022-01-05 Author has 620 answers
\(\displaystyle{a}_{{n}}={\frac{{{8}{n}^{{2}}+{1}{n}+{1}}}{{{7}{n}^{{2}}+{9}{n}+{9}}}}\)
\(\displaystyle\lim_{{{n}\rightarrow\infty}}{a}_{{n}}=\lim_{{{n}\rightarrow\infty}}{\frac{{{b}{n}^{{{2}}}+{n}+{1}}}{{{7}{n}^{{{2}}}+{9}{n}+{9}}}}\)
\(\displaystyle\lim_{{{n}\rightarrow\infty}}{\frac{{{n}^{{{2}}}{\left({b}+{\frac{{{1}}}{{{n}}}}+{\frac{{{1}}}{{{n}^{{{2}}}}}}\right)}}}{{{n}^{{{2}}}{\left({7}+{\frac{{{9}}}{{{n}}}}+{\frac{{{9}}}{{{n}^{{{2}}}}}}\right)}}}}\)
\(\displaystyle\lim_{{{n}\rightarrow\infty}}{a}_{{n}}=\lim_{{{n}\rightarrow\infty}}{\frac{{{b}+{\frac{{{1}}}{{{n}}}}+{\frac{{{1}}}{{{n}^{{{2}}}}}}}}{{{7}+{\frac{{{9}}}{{{n}}}}+{\frac{{{9}}}{{{n}^{{{2}}}}}}}}}\)
\(\displaystyle\Rightarrow\lim_{{{n}\rightarrow\infty}}{a}_{{n}}={\frac{{{b}+{1}\times{0}+{1}\times{0}}}{{{7}+{9}\times{0}+{9}\times{0}}}}={\frac{{{b}+{0}+{0}}}{{{7}+{0}+{0}}}}={\frac{{{b}}}{{{7}}}}\)
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