What is the limit as x approaches infinity of sqrt((12x^3-4x+1)/(1+3x+2x^3)) ?

The limit as x approaches infinity of sqrt((12x^3-4x+1)/(1+3x+2x^3)) is square root of 6

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limit as x approaches infinity of sqrt((12x^3-4x+1)/(1+3x+2x^3))

x \to \infty lim (\frac{12 x ^{3} - 4 x + 1}{1 + 3 x + 2 x ^{3}})

6

Decimal Notation

2.44948 \dots

Solution steps

x \to \infty lim (\frac{12 x ^{3} - 4 x + 1}{1 + 3 x + 2 x ^{3}})

Divide by highest denominator power:

\frac{12 - \frac{4}{x ^{2}} + \frac{1}{x ^{3}}}{\frac{1}{x ^{3}} + \frac{3}{x ^{2}} + 2}

= x \to \infty lim (\frac{12 - \frac{4}{x ^{2}} + \frac{1}{x ^{3}}}{\frac{1}{x ^{3}} + \frac{3}{x ^{2}} + 2})

x \to a lim [f (x)]^{b} = [x \to a lim f (x)]^{b}

With the exception of indeterminate form

= x \to \infty lim (\frac{12 - \frac{4}{x ^{2}} + \frac{1}{x ^{3}}}{\frac{1}{x ^{3}} + \frac{3}{x ^{2}} + 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} ( 12 - \frac{4}{x ^{2}} + \frac{1}{x ^{3}} )}{lim _{x \to \infty} ( \frac{1}{x ^{3}} + \frac{3}{x ^{2}} + 2 )}

x \to \infty lim (12 - \frac{4}{x ^{2}} + \frac{1}{x ^{3}}) = 12

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

= \frac{12}{2}

Simplify

= 6

\frac{8 x - 7}{5 x + 12}

\frac{d}{d x} (f (x) (x)^{- 1} (a x))

\frac{d}{d t} (t^{2} - 1)

\int \frac{sin ( x )}{2 e ^{x}} d x

\int_{- 3}^{3} ∣ x ∣ - (x^{2} - 6) d x

What is the limit as x approaches infinity of sqrt((12x^3-4x+1)/(1+3x+2x^3)) ?
The limit as x approaches infinity of sqrt((12x^3-4x+1)/(1+3x+2x^3)) is square root of 6