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

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

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

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

- \frac{1}{6}

Solution steps

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

Multiply by the conjugate of

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

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

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

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

Divide by highest denominator power:

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

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

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

x \to \infty lim (1) = 1

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

= - \frac{1}{6}

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\frac{d}{d x} (\frac{e ^{2 x}}{e ^{2 x} + 1})

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