What is the limit as x approaches infinity of (4x^3+2x-24)/(x^4-x^2+84) ?

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limit as x approaches infinity of (4x^3+2x-24)/(x^4-x^2+84)

x \to \infty lim (\frac{4 x ^{3} + 2 x - 24}{x ^{4} - x ^{2} + 84})

0

Solution steps

x \to \infty lim (\frac{4 x ^{3} + 2 x - 24}{x ^{4} - x ^{2} + 84})

Divide by highest denominator power:

\frac{\frac{4}{x} + \frac{2}{x ^{3}} - \frac{24}{x ^{4}}}{1 - \frac{1}{x ^{2}} + \frac{84}{x ^{4}}}

= x \to \infty lim (\frac{\frac{4}{x} + \frac{2}{x ^{3}} - \frac{24}{x ^{4}}}{1 - \frac{1}{x ^{2}} + \frac{84}{x ^{4}}})

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{4}{x} + \frac{2}{x ^{3}} - \frac{24}{x ^{4}} )}{lim _{x \to \infty} ( 1 - \frac{1}{x ^{2}} + \frac{84}{x ^{4}} )}

x \to \infty lim (\frac{4}{x} + \frac{2}{x ^{3}} - \frac{24}{x ^{4}}) = 0

x \to \infty lim (1 - \frac{1}{x ^{2}} + \frac{84}{x ^{4}}) = 1

= \frac{0}{1}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

x \to 3 lim (\frac{2 x}{x - 2})

x \to 3 lim (\frac{5 - 2 x}{x - 3})

x \to 7 - lim (\frac{x - x}{x - 7})

x \to 5 lim (2 x^{2} - 1)

x \to 1 lim (\frac{x - 1}{x ^{2} + x})

What is the limit as x approaches infinity of (4x^3+2x-24)/(x^4-x^2+84) ?
The limit as x approaches infinity of (4x^3+2x-24)/(x^4-x^2+84) is 0