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

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

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limit as h approaches infinity of (5h^4-2h^2+3)/(3h^3+2h^2+h)

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

\infty

Solution steps

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

Divide by highest denominator power:

\frac{5 h - \frac{2}{h} + \frac{3}{h ^{3}}}{3 + \frac{2}{h} + \frac{1}{h ^{2}}}

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

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

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

= \frac{\infty}{3}

Apply Infinity Property:

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

= \infty

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

\frac{d y}{d x} = 2 x (1 + y) e^{x^{2}}

y = 2 x^{2} + 1

\frac{d}{d x} (\frac{( x + 1 ) ^{3}}{3} + \frac{1}{4 x + 4})

3 x^{2} e^{x}

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