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

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

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

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

\infty

Solution steps

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

Divide by highest denominator power:

\frac{2 x - \frac{5}{x} + \frac{4}{x ^{2}}}{1 + \frac{3}{x} - \frac{4}{x ^{2}}}

= x \to \infty lim (\frac{2 x - \frac{5}{x} + \frac{4}{x ^{2}}}{1 + \frac{3}{x} - \frac{4}{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} ( 2 x - \frac{5}{x} + \frac{4}{x ^{2}} )}{lim _{x \to \infty} ( 1 + \frac{3}{x} - \frac{4}{x ^{2}} )}

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

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

= \frac{\infty}{1}

Apply Infinity Property:

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

= \infty

x cos (x)

\int \frac{9}{x ^{2} - 6 x + 18} d x

\frac{0.6}{s}

\int_{- 1}^{0} - 1 d x

\int cos (\frac{nπ x}{L}) sin (\frac{nπ x}{L}) d x

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