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

The limit as x approaches infinity of (x^2-6)/(2+x-3x^2) is -1/3

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

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

- \frac{1}{3}

Decimal

- 0.33333 \dots

Solution steps

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

Divide by highest denominator power:

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

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

x \to \infty lim (1 - \frac{6}{x ^{2}}) = 1

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

= \frac{1}{- 3}

Simplify

= - \frac{1}{3}

x \to \infty lim (\frac{16 x ^{4} + 64 x ^{2} + x ^{2}}{2 x ^{2} - 4})

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

x \to 4 + lim (\frac{x - 1}{4 - x})

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

x \to π lim (tan (\frac{5 x}{6}))

What is the limit as x approaches infinity of (x^2-6)/(2+x-3x^2) ?
The limit as x approaches infinity of (x^2-6)/(2+x-3x^2) is -1/3