# What is the limit of  [x^2-4x+4]/[x^3-64] as x goes to infinity?

Oct 11, 2015

It is $0$.

#### Explanation:

${\lim}_{x \rightarrow \infty} \frac{{x}^{2} - 4 x + 4}{{x}^{3} - 64}$ has indeterminate form $\frac{\infty}{\infty}$

Make the denominator not go to $\infty$:

lim_(xrarroo) [x^2-4x+4]/[x^3-64] = lim_(xrarroo) [x^3(1/x-4/x^2+4/x^3)]/[x^3(1-64/x^3)

$= {\lim}_{x \rightarrow \infty} \frac{\frac{1}{x} - \frac{4}{x} ^ 2 + \frac{4}{x} ^ 3}{1 - \frac{64}{x} ^ 3}$

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