What is the limit as x approaches infinity of 2x-sqrt(4x^2-2x) ?

The limit as x approaches infinity of 2x-sqrt(4x^2-2x) is 1/2

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 2x-sqrt(4x^2-2x)

x \to \infty lim (2 x - 4 x^{2} - 2 x)

\frac{1}{2}

Decimal Notation

0.5

Solution steps

x \to \infty lim (2 x - 4 x^{2} - 2 x)

Multiply by the conjugate of

2 x - 4 x^{2} - 2 x : \frac{2 x}{2 x + 4 x ^{2} - 2 x}

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

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

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

Divide by highest denominator power:

\frac{1}{2 + 4 - \frac{2}{x}}

= 2 \cdot x \to \infty lim \frac{1}{2 + 4 - \frac{2}{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

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

x \to \infty lim (1) = 1

x \to \infty lim (2 + 4 - \frac{2}{x}) = 4

= 2 \cdot \frac{1}{4}

Simplify

2 \cdot \frac{1}{4} : \frac{1}{2}

= \frac{1}{2}

\int \frac{1}{2 x + 2} d x

5^{t}

\int 2 z ln (y) d z

\int \frac{15}{x} d x

\int 6 e^{5 x} d x

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