What is the limit as n approaches infinity of (3n^2+7n+11)/(8n^2-5n+3) ?

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limit as n approaches infinity of (3n^2+7n+11)/(8n^2-5n+3)

n \to \infty lim (\frac{3 n ^{2} + 7 n + 11}{8 n ^{2} - 5 n + 3})

\frac{3}{8}

Decimal Notation

0.375

Solution steps

n \to \infty lim (\frac{3 n ^{2} + 7 n + 11}{8 n ^{2} - 5 n + 3})

Divide by highest denominator power:

\frac{3 + \frac{7}{n} + \frac{11}{n ^{2}}}{8 - \frac{5}{n} + \frac{3}{n ^{2}}}

= n \to \infty lim (\frac{3 + \frac{7}{n} + \frac{11}{n ^{2}}}{8 - \frac{5}{n} + \frac{3}{n ^{2}}})

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

= \frac{lim _{n \to \infty} ( 3 + \frac{7}{n} + \frac{11}{n ^{2}} )}{lim _{n \to \infty} ( 8 - \frac{5}{n} + \frac{3}{n ^{2}} )}

n \to \infty lim (3 + \frac{7}{n} + \frac{11}{n ^{2}}) = 3

n \to \infty lim (8 - \frac{5}{n} + \frac{3}{n ^{2}}) = 8

= \frac{3}{8}

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What is the limit as n approaches infinity of (3n^2+7n+11)/(8n^2-5n+3) ?
The limit as n approaches infinity of (3n^2+7n+11)/(8n^2-5n+3) is 3/8