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

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

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 (6x^2+2x)/(2x^2-2x+1)

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

3

Solution steps

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

Divide by highest denominator power:

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

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

x \to \infty lim (6 + \frac{2}{x}) = 6

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

= \frac{6}{2}

Simplify

= 3

\int cos^{3} (\frac{x}{2}) d x

\frac{π r ^{2} ( π r ^{2} - 1 )}{4}

\frac{\partial}{\partial x} (x y e^{- x^{2} - y^{2}})

f (x) = (6 x - 8)^{\frac{1}{2}}, at x = 4

\int_{0}^{π} x^{2} cos (n x) d x

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