What is the limit as n approaches infinity of (n^2+3)/(n^3sqrt(n)+n^{5/2)} ?

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limit as n approaches infinity of (n^2+3)/(n^3sqrt(n)+n^{5/2)}

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

0

Solution steps

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

Divide by highest denominator power:

\frac{\frac{1}{n} + \frac{3}{n ^{3}}}{n ^{\frac{1}{2}} + \frac{1}{n ^{\frac{1}{2}}}}

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

n \to \infty lim (\frac{1}{n} + \frac{3}{n ^{3}}) = 0

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

= \frac{0}{\infty}

Apply Infinity Property:

\frac{0}{\infty} = 0

= 0

\int \frac{x ^{3} + 3 x - 2}{x ^{2} - x} d x

\frac{d}{d x} (\frac{7}{x ^{7}})

\int_{1}^{8} x^{- \frac{8}{3}} d x

y = x^{2} + 9, y = 10 + 2 x

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

What is the limit as n approaches infinity of (n^2+3)/(n^3sqrt(n)+n^{5/2)} ?
The limit as n approaches infinity of (n^2+3)/(n^3sqrt(n)+n^{5/2)} is 0