What is the limit as x approaches-infinity of x/(9-x^2) ?

The limit as x approaches-infinity of x/(9-x^2) is 0

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limit as x approaches-infinity of x/(9-x^2)

x \to - \infty lim (\frac{x}{9 - x ^{2}})

0

Solution steps

x \to - \infty lim (\frac{x}{9 - x ^{2}})

Divide by highest denominator power:

\frac{\frac{1}{x}}{\frac{9}{x ^{2}} - 1}

= x \to - \infty lim (\frac{\frac{1}{x}}{\frac{9}{x ^{2}} - 1})

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} )}{lim _{x \to - \infty} ( \frac{9}{x ^{2}} - 1 )}

x \to - \infty lim (\frac{1}{x}) = 0

x \to - \infty lim (\frac{9}{x ^{2}} - 1) = - 1

= \frac{0}{- 1}

Apply rule

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

= 0

x^{2} y^{^{''}} + 2 x y^{^{'}} - 72 y = 0

\int_{0}^{2} \frac{x + 1}{x ( x ^{2} + 1 )} d x

y^{^{'}} = \frac{( x - y sin ( 2 x ) )}{sin ^{2} ( x )}

n = 1 \sum \infty \frac{1}{n \cdot 3 ^{n}}

3 e^{- 7 x}

What is the limit as x approaches-infinity of x/(9-x^2) ?
The limit as x approaches-infinity of x/(9-x^2) is 0