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

The limit as x approaches infinity of 3x-sqrt(9x^2+6x) is -1

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limit as x approaches infinity of 3x-sqrt(9x^2+6x)

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

- 1

Solution steps

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

Multiply by the conjugate of

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

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

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

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

Divide by highest denominator power:

\frac{1}{3 + 9 + \frac{6}{x}}

= - 6 \cdot x \to \infty lim \frac{1}{3 + 9 + \frac{6}{x}}

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

= - 6 \cdot \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( 3 + 9 + \frac{6}{x} )}

x \to \infty lim (1) = 1

x \to \infty lim (3 + 9 + \frac{6}{x}) = 6

= - 6 \cdot \frac{1}{6}

Simplify

- 6 \cdot \frac{1}{6} : - 1

= - 1

\int_{0}^{1} \frac{x}{1 + x ^{4}} d x

\frac{d y}{d x} = x y + 2 x + y + 2

y = 3 x, [3, 7]

\int 34 arctan (x) d x

x \to 2 lim ((x + 2) \cdot (2 x - 1))

What is the limit as x approaches infinity of 3x-sqrt(9x^2+6x) ?
The limit as x approaches infinity of 3x-sqrt(9x^2+6x) is -1