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

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Answer 1 · 2015-10-11T03:10:52

It is $0$ .

$lim_(xrarroo) [x^2-4x+4]/[x^3-64]$ has indeterminate form $oo/oo$

Make the denominator not go to $oo$ :

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

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

$= (0-0+0)/(1-0) = 0$

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