# How do you find the limit of (sqrt(x+6)-x)/(x^3-3x^2) as x->-oo?

Feb 6, 2018

$0$

#### Explanation:

$\frac{\sqrt{x + 6} - x}{{x}^{3} - 3 {x}^{2}}$

Divide by ${x}^{3}$:

$\frac{\frac{\sqrt{x + 6}}{x} ^ 3 - \frac{x}{x} ^ 3}{{x}^{3} / {x}^{3} - 3 {x}^{2} / {x}^{3}} = \frac{\frac{\sqrt{x + 6}}{x} ^ 3 - \frac{1}{x} ^ 2}{1 - \frac{3}{x}}$

lim_(x->-oo)((sqrt(x+6))/x^3-1/x^2)/(1-3/x)=(lim_(x->-oo)((sqrt(x+6))/x^3-1/x^2))/(lim_(x->-oo)(1-3/x)

${\lim}_{x \to - \infty} \left(\frac{\sqrt{x + 6}}{x} ^ 3 - \frac{1}{x} ^ 2\right) = {\lim}_{x \to - \infty} \frac{\sqrt{x + 6}}{x} ^ 3 - {\lim}_{x \to - \infty} \left(\frac{1}{x} ^ 2\right)$

${\lim}_{x \to - \infty} \frac{\sqrt{x + 6}}{x} ^ 3 = 0$

${\lim}_{x \to - \infty} \left(\frac{1}{x} ^ 2\right) = 0$

$0 - 0 = 0$

${\lim}_{x \to - \infty} \left(1 - \frac{3}{x}\right) = {\lim}_{x \to - \infty} \left(1\right) - {\lim}_{x \to - \infty} \left(\frac{3}{x}\right)$

${\lim}_{x \to - \infty} \left(1\right) = 1$

${\lim}_{x \to - \infty} \left(\frac{3}{x}\right) = 0$

$\therefore$

$1 - 0 = 1$

$\frac{0}{1} = 0$

Hence:

${\lim}_{x \to - \infty} \frac{\sqrt{x + 6} - x}{{x}^{3} - 3 {x}^{2}} = 0$