How do you find the limit of $\frac{x^{2} + x - 12}{x - 3}$ $(x^2+x-12)/(x-3)$ as $x \to - 4$ $x->-4$ ?

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Answer 1 · 2017-02-20T19:49:42

Substitution.

$lim_(xrarr-4)(x^2+x-12) = 16+(-4)-12 = 0$

$lim_(xrarr-4)(x-3) = -7$

So,

$lim_(xrarr-4)(x^2+x-12)/(x-3) = 0/(-7) = 0$

How do you find the limit of (x^2+x-12)/(x-3)x2+x−12x−3(x^2+x-12)/(x-3) as x->-4x→−4x->-4?