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

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

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

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

- \infty

Solution steps

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

Divide by highest denominator power:

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

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

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

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

= \frac{- \infty}{1}

Apply Infinity Property:

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

= - \infty

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

\int \frac{3 cos ( ln ( x ) )}{x} d x

\frac{d}{d x} (arctan (4^{x}))

\frac{d}{d x} (ln (x - 2) + \frac{x - 3}{x - 2})

6 x + 1

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