What is the limit as x approaches infinity of ((6x^3-5x+7))/((3+x-3x^2-2x^3)) ?

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limit as x approaches infinity of ((6x^3-5x+7))/((3+x-3x^2-2x^3))

x \to \infty lim (\frac{( 6 x ^{3} - 5 x + 7 )}{( 3 + x - 3 x ^{2} - 2 x ^{3} )})

- 3

Solution steps

x \to \infty lim (\frac{6 x ^{3} - 5 x + 7}{3 + x - 3 x ^{2} - 2 x ^{3}})

Divide by highest denominator power:

\frac{6 - \frac{5}{x ^{2}} + \frac{7}{x ^{3}}}{\frac{3}{x ^{3}} + \frac{1}{x ^{2}} - \frac{3}{x} - 2}

= x \to \infty lim (\frac{6 - \frac{5}{x ^{2}} + \frac{7}{x ^{3}}}{\frac{3}{x ^{3}} + \frac{1}{x ^{2}} - \frac{3}{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} ( 6 - \frac{5}{x ^{2}} + \frac{7}{x ^{3}} )}{lim _{x \to \infty} ( \frac{3}{x ^{3}} + \frac{1}{x ^{2}} - \frac{3}{x} - 2 )}

x \to \infty lim (6 - \frac{5}{x ^{2}} + \frac{7}{x ^{3}}) = 6

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

= \frac{6}{- 2}

Simplify

= - 3

x \to 1 lim (x ln ∣ x - 1 ∣)

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

x \to - 1 lim (5 x)

x \to 1 + lim (x^{\frac{8}{( x - 1 )}})

x \to \infty lim (\frac{( x ^{13} )}{e ^{x}})

What is the limit as x approaches infinity of ((6x^3-5x+7))/((3+x-3x^2-2x^3)) ?
The limit as x approaches infinity of ((6x^3-5x+7))/((3+x-3x^2-2x^3)) is -3