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

The limit as x approaches infinity of (2x^3+3x-4)/(5-4x^3) is -1/2

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

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

- \frac{1}{2}

Decimal Notation

- 0.5

Solution steps

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

Divide by highest denominator power:

\frac{2 + \frac{3}{x ^{2}} - \frac{4}{x ^{3}}}{\frac{5}{x ^{3}} - 4}

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

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

x \to \infty lim (\frac{5}{x ^{3}} - 4) = - 4

= \frac{2}{- 4}

Simplify

\frac{2}{- 4} : - \frac{1}{2}

= - \frac{1}{2}

f^{^{'}} (x) = x^{2} e^{x}

\frac{d}{d x} (\frac{6 x ^{\frac{3}{2}}}{x})

x \to \frac{1}{4} lim (x^{2} - x + \frac{9}{16})

\int 1 + 81 x d x

\int \frac{x ^{2}}{16} d x

What is the limit as x approaches infinity of (2x^3+3x-4)/(5-4x^3) ?
The limit as x approaches infinity of (2x^3+3x-4)/(5-4x^3) is -1/2