What is the limit as x approaches infinity of (10x^3-7x+2)/(2x^2+9x+6x^3) ?

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limit as x approaches infinity of (10x^3-7x+2)/(2x^2+9x+6x^3)

x \to \infty lim (\frac{10 x ^{3} - 7 x + 2}{2 x ^{2} + 9 x + 6 x ^{3}})

\frac{5}{3}

Decimal

1.66666 \dots

Solution steps

x \to \infty lim (\frac{10 x ^{3} - 7 x + 2}{2 x ^{2} + 9 x + 6 x ^{3}})

Divide by highest denominator power:

\frac{10 - \frac{7}{x ^{2}} + \frac{2}{x ^{3}}}{\frac{2}{x} + \frac{9}{x ^{2}} + 6}

= x \to \infty lim (\frac{10 - \frac{7}{x ^{2}} + \frac{2}{x ^{3}}}{\frac{2}{x} + \frac{9}{x ^{2}} + 6})

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} ( 10 - \frac{7}{x ^{2}} + \frac{2}{x ^{3}} )}{lim _{x \to \infty} ( \frac{2}{x} + \frac{9}{x ^{2}} + 6 )}

x \to \infty lim (10 - \frac{7}{x ^{2}} + \frac{2}{x ^{3}}) = 10

x \to \infty lim (\frac{2}{x} + \frac{9}{x ^{2}} + 6) = 6

= \frac{10}{6}

Cancel the common factor:

2

= \frac{5}{3}

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

x \to 0 lim (- \frac{2}{x ^{4}})

x \to 0 lim (\frac{7}{x ^{2}})

x \to \infty lim (in (x))

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

What is the limit as x approaches infinity of (10x^3-7x+2)/(2x^2+9x+6x^3) ?
The limit as x approaches infinity of (10x^3-7x+2)/(2x^2+9x+6x^3) is 5/3