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

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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Raymond Foley
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 $$\displaystyle{n}\rightarrow\infty$$, $$\displaystyle{\frac{{{1}}}{{{n}}}}\rightarrow{0}$$
$$\displaystyle{n}\rightarrow\infty$$, $$\displaystyle{\frac{{{1}}}{{{n}^{{{2}}}}}}\rightarrow{0}$$
$$\displaystyle{n}\rightarrow\infty$$, $$\displaystyle{\frac{{{9}}}{{{n}}}}\rightarrow{0}$$
$$\displaystyle{n}\rightarrow\infty$$, $$\displaystyle{\frac{{{9}}}{{{n}^{{{2}}}}}}\rightarrow{0}$$
$$\displaystyle\lim_{{{n}\rightarrow\infty}}{a}_{{n}}={\frac{{{8}+{0}+{0}}}{{{7}+{0}+{0}}}}$$
$$\displaystyle={\frac{{{8}}}{{{7}}}}$$