What is the limit as x approaches infinity of sqrt(x^2+12x)-sqrt(x^2+6x) ?

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limit as x approaches infinity of sqrt(x^2+12x)-sqrt(x^2+6x)

x \to \infty lim (x^{2} + 12 x - x^{2} + 6 x)

3

Solution steps

x \to \infty lim (x^{2} + 12 x - x^{2} + 6 x)

Multiply by the conjugate of

x^{2} + 12 x - x^{2} + 6 x : \frac{6 x}{x ^{2} + 12 x + x ^{2} + 6 x}

= x \to \infty lim (\frac{6 x}{x ^{2} + 12 x + x ^{2} + 6 x})

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

= 6 \cdot x \to \infty lim (\frac{x}{x ^{2} + 12 x + x ^{2} + 6 x})

Divide by

x^{2} + 12 x : \frac{\frac{x}{x ^{2} + 12 x}}{1 + \frac{x + 6}{x + 12}}

= 6 \cdot x \to \infty lim \frac{\frac{x}{x ^{2} + 12 x}}{1 + \frac{x + 6}{x + 12}}

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

= 6 \cdot \frac{lim _{x \to \infty} ( \frac{x}{x ^{2} + 12 x} )}{lim _{x \to \infty} ( 1 + \frac{x + 6}{x + 12} )}

x \to \infty lim (\frac{x}{x ^{2} + 12 x}) = 1

x \to \infty lim (1 + \frac{x + 6}{x + 12}) = 2

= 6 \cdot \frac{1}{2}

Simplify

6 \cdot \frac{1}{2} : 3

= 3

x \to 0 lim (x csc (5 x))

x \to - 1 - lim (x^{2} + a)

x \to \infty lim (23)

x \to 0 lim (\frac{3 27 x ^{3} - 2 x + 5}{3 x - 1})

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

What is the limit as x approaches infinity of sqrt(x^2+12x)-sqrt(x^2+6x) ?
The limit as x approaches infinity of sqrt(x^2+12x)-sqrt(x^2+6x) is 3