What is the limit as x approaches infinity of sqrt(9x^2-18x+1)-3x ?

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as x approaches infinity of sqrt(9x^2-18x+1)-3x

x \to \infty lim (9 x^{2} - 18 x + 1 - 3 x)

- 3

Solution steps

x \to \infty lim (9 x^{2} - 18 x + 1 - 3 x)

Multiply by the conjugate of

9 x^{2} - 18 x + 1 - 3 x : \frac{- 18 x + 1}{9 x ^{2} - 18 x + 1 + 3 x}

= x \to \infty lim (\frac{- 18 x + 1}{9 x ^{2} - 18 x + 1 + 3 x})

Divide by highest denominator power:

\frac{- 18 + \frac{1}{x}}{9 - \frac{18}{x} + \frac{1}{x ^{2}} + 3}

= x \to \infty lim \frac{- 18 + \frac{1}{x}}{9 - \frac{18}{x} + \frac{1}{x ^{2}} + 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} ( - 18 + \frac{1}{x} )}{lim _{x \to \infty} ( 9 - \frac{18}{x} + \frac{1}{x ^{2}} + 3 )}

x \to \infty lim (- 18 + \frac{1}{x}) = - 18

x \to \infty lim (9 - \frac{18}{x} + \frac{1}{x ^{2}} + 3) = 6

= \frac{- 18}{6}

Simplify

= - 3

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

x \to 0 lim (\frac{- 1}{∣ x ∣})

x \to 0 lim ((1 + 2 x)^{\frac{2}{x}})

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

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

What is the limit as x approaches infinity of sqrt(9x^2-18x+1)-3x ?
The limit as x approaches infinity of sqrt(9x^2-18x+1)-3x is -3