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

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

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Popular Calculus >

limit as x approaches infinity of x(x-sqrt(x^2-4))

Solution

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

Solution

2

Solution steps

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

Multiply by the conjugate of

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

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

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

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

Divide by highest denominator power:

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

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

Simplify

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

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

Apply L'Hopital's Rule

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

Simplify

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

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

Apply L'Hopital's Rule

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

Simplify

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

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

Apply L'Hopital's Rule

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

Simplify

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

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

Apply L'Hopital's Rule

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

Simplify

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

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

Apply L'Hopital's Rule

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

Simplify

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

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

Apply L'Hopital's Rule

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

Simplify

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

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

Apply L'Hopital's Rule

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

Simplify

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

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

Apply L'Hopital's Rule

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

Simplify

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

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

Divide by highest denominator power:

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

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

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

x \to \infty lim (1) = 1

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

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

Simplify

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

= 2

Graph

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Frequently Asked Questions (FAQ)

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