What is the limit as x approaches infinity of (9x-x^2-7x^4)/(x^3+12x) ?

The limit as x approaches infinity of (9x-x^2-7x^4)/(x^3+12x) is -infinity

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limit as x approaches infinity of (9x-x^2-7x^4)/(x^3+12x)

x \to \infty lim (\frac{9 x - x ^{2} - 7 x ^{4}}{x ^{3} + 12 x})

- \infty

Solution steps

x \to \infty lim (\frac{9 x - x ^{2} - 7 x ^{4}}{x ^{3} + 12 x})

Divide by highest denominator power:

\frac{\frac{9}{x ^{2}} - \frac{1}{x} - 7 x}{1 + \frac{12}{x ^{2}}}

= x \to \infty lim (\frac{\frac{9}{x ^{2}} - \frac{1}{x} - 7 x}{1 + \frac{12}{x ^{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 _{x \to \infty} ( \frac{9}{x ^{2}} - \frac{1}{x} - 7 x )}{lim _{x \to \infty} ( 1 + \frac{12}{x ^{2}} )}

x \to \infty lim (\frac{9}{x ^{2}} - \frac{1}{x} - 7 x) = - \infty

x \to \infty lim (1 + \frac{12}{x ^{2}}) = 1

= \frac{- \infty}{1}

Apply Infinity Property:

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

= - \infty

\int \frac{( x ^{6} e ^{x} )}{x} d x

\int (1 - x) x^{3} (x - 2)^{2} d x

x y^{^{'}} - 2 y = 2 x^{2} - 2 x - 2

\frac{d}{d x} (5 x^{0.5})

6 x + 9

What is the limit as x approaches infinity of (9x-x^2-7x^4)/(x^3+12x) ?
The limit as x approaches infinity of (9x-x^2-7x^4)/(x^3+12x) is -infinity