What is the limit as x approaches infinity of (9x^3+x^2-5)/(3x^4+4x^2+3) ?

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limit as x approaches infinity of (9x^3+x^2-5)/(3x^4+4x^2+3)

x \to \infty lim (\frac{9 x ^{3} + x ^{2} - 5}{3 x ^{4} + 4 x ^{2} + 3})

0

Solution steps

x \to \infty lim (\frac{9 x ^{3} + x ^{2} - 5}{3 x ^{4} + 4 x ^{2} + 3})

Divide by highest denominator power:

\frac{\frac{9}{x} + \frac{1}{x ^{2}} - \frac{5}{x ^{4}}}{3 + \frac{4}{x ^{2}} + \frac{3}{x ^{4}}}

= x \to \infty lim (\frac{\frac{9}{x} + \frac{1}{x ^{2}} - \frac{5}{x ^{4}}}{3 + \frac{4}{x ^{2}} + \frac{3}{x ^{4}}})

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

x \to \infty lim (\frac{9}{x} + \frac{1}{x ^{2}} - \frac{5}{x ^{4}}) = 0

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

= \frac{0}{3}

Apply rule

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

= 0

x \to 15 lim (x)

x \to - 3 lim (7 x + 1)

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

x \to - 5 lim (\frac{3 x}{2 x + 10})

x \to 0 lim (\frac{\frac{1}{x}}{x})

What is the limit as x approaches infinity of (9x^3+x^2-5)/(3x^4+4x^2+3) ?
The limit as x approaches infinity of (9x^3+x^2-5)/(3x^4+4x^2+3) is 0