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

The limit as x approaches-infinity of (x^2+2)/(x^3+x^2-1) is 0

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

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

0

Solution steps

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

Divide by highest denominator power:

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

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

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

x \to - \infty lim (\frac{1}{x} + \frac{2}{x ^{3}}) = 0

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

= \frac{0}{1}

Apply rule

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

= 0

x \to 0 lim (7 x^{3} + \frac{1}{4} x^{2})

\frac{\partial}{\partial y} (x \cdot ln (x + 2 y) + \frac{x}{y})

f (x) = \frac{1}{x + 2}

x \to 0 lim (2 x (1 - x))

\int_{1}^{\infty} x^{- \frac{1}{2}} d x

What is the limit as x approaches-infinity of (x^2+2)/(x^3+x^2-1) ?
The limit as x approaches-infinity of (x^2+2)/(x^3+x^2-1) is 0