What is the limit as n approaches infinity of (2n^5-3n^4+8)/(3n^5+1) ?

The limit as n approaches infinity of (2n^5-3n^4+8)/(3n^5+1) is 2/3

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limit as n approaches infinity of (2n^5-3n^4+8)/(3n^5+1)

n \to \infty lim (\frac{2 n ^{5} - 3 n ^{4} + 8}{3 n ^{5} + 1})

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

n \to \infty lim (\frac{2 n ^{5} - 3 n ^{4} + 8}{3 n ^{5} + 1})

Divide by highest denominator power:

\frac{2 - \frac{3}{n} + \frac{8}{n ^{5}}}{3 + \frac{1}{n ^{5}}}

= n \to \infty lim (\frac{2 - \frac{3}{n} + \frac{8}{n ^{5}}}{3 + \frac{1}{n ^{5}}})

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} ( 2 - \frac{3}{n} + \frac{8}{n ^{5}} )}{lim _{n \to \infty} ( 3 + \frac{1}{n ^{5}} )}

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

n \to \infty lim (3 + \frac{1}{n ^{5}}) = 3

= \frac{2}{3}

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What is the limit as n approaches infinity of (2n^5-3n^4+8)/(3n^5+1) ?
The limit as n approaches infinity of (2n^5-3n^4+8)/(3n^5+1) is 2/3