What is the limit as n approaches infinity of (\sqrt[3]{2n^7+7n^3+1})/(3n^2-1) ?

The limit as n approaches infinity of (\sqrt[3]{2n^7+7n^3+1})/(3n^2-1) is infinity

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limit as n approaches infinity of (\sqrt[3]{2n^7+7n^3+1})/(3n^2-1)

n \to \infty lim (\frac{3 2 n ^{7} + 7 n ^{3} + 1}{3 n ^{2} - 1})

\infty

Solution steps

n \to \infty lim (\frac{3 2 n ^{7} + 7 n ^{3} + 1}{3 n ^{2} - 1})

Divide by highest denominator power:

\frac{n ^{\frac{1}{3}} 3 \frac{7}{n ^{4}} + \frac{1}{n ^{7}} + 2}{3 - \frac{1}{n ^{2}}}

= n \to \infty lim \frac{n ^{\frac{1}{3}} 3 \frac{7}{n ^{4}} + \frac{1}{n ^{7}} + 2}{3 - \frac{1}{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} ( n ^{\frac{1}{3}} 3 \frac{7}{n ^{4}} + \frac{1}{n ^{7}} + 2 )}{lim _{n \to \infty} ( 3 - \frac{1}{n ^{2}} )}

n \to \infty lim (n^{\frac{1}{3}} 3 \frac{7}{n ^{4}} + \frac{1}{n ^{7}} + 2) = \infty

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

= \frac{\infty}{3}

Apply Infinity Property:

\frac{\infty}{c} = \infty

= \infty

x \to 2 lim (\frac{∣ x - 1 ∣}{x - 1})

x \to 0 lim (x \cdot e^{- x})

x \to 0 lim (\frac{x}{( x + 3 ) ^{2} - 9})

x \to \frac{3 π}{4} lim (2 cos (x))

x \to 101 lim (\frac{x}{x + 4})

What is the limit as n approaches infinity of (\sqrt[3]{2n^7+7n^3+1})/(3n^2-1) ?
The limit as n approaches infinity of (\sqrt[3]{2n^7+7n^3+1})/(3n^2-1) is infinity