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

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

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limit as x approaches-infinity of (x^3+x^2-3x+2)/(x^2-3x+1)

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

- \infty

Solution steps

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

Divide by highest denominator power:

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

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

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

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

= \frac{- \infty}{1}

Apply Infinity Property:

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

= - \infty

\frac{d}{d x} (- \frac{1}{2} sin^{2} (x))

2 x^{2} - x

\int \frac{7 x ^{2} - 6 x + 8}{x ^{3} + 4 x} d x

\frac{d}{d x} (x sin (x y))

t \to \infty lim (cos (t))

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