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

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

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

- 2

Solution steps

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

Simplify

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

Multiply by the conjugate of

x^{2} + 2 x - x^{2} + 6 x : \frac{- 4 x}{x ^{2} + 2 x + x ^{2} + 6 x}

= x \to \infty lim (\frac{- 4 x}{x ^{2} + 2 x + x ^{2} + 6 x})

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

= - 4 \cdot x \to \infty lim (\frac{x}{x ^{2} + 2 x + x ^{2} + 6 x})

Divide by highest denominator power:

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

= - 4 \cdot x \to \infty lim \frac{1}{1 + \frac{2}{x} + 1 + \frac{6}{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{2}{x} + 1 + \frac{6}{x} )}

x \to \infty lim (1) = 1

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

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

Simplify

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

= - 2

x \to \infty lim (5 x - 2)

h \to 0 lim (\frac{3 h}{h})

x \to \infty lim (\frac{3 x + 1}{2 x + 1})

x \to 7 - lim (\frac{2}{x - 7})

x \to \infty lim (5 x + 3)

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