How do you use the epsilon delta definition to prove that the limit of $x^{3} + 6 x^{2} = 32$ $x^3+6x^2=32$ as $x \to 2$ $x->2$ ?

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Answer 1 · 2016-10-26T15:39:46

$delta(epsilon)=min(1;epsilon/49)$

$abs(x^3+6x^2-32)=abs(x-2)(x+4)^2$
$abs(x-2) < min(1;epsilon/49) rArr (x+4)^2<(2+1+4)^2=49$
and
$abs(x^3+6x^2-32) < epsilon/49 * 49 = epsilon$

How do you use the epsilon delta definition to prove that the limit of x^3+6x^2=32x3+6x2=32x^3+6x^2=32 as x->2x→2x->2?