How do you find the limit of $f (x) = \frac{x^{2} + x - 6}{x + 3}$ $f(x) = (x^2 + x - 6) / (x + 3)$ as x approaches 0?

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Answer 1 · 2016-02-18T18:23:57

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$lim_(x rarr 0) (x^2+x-6)/(x+3)=(0^2+0-6)/(0+3)=-6/3=-2$

There is no indetermination. Maybe you have made a mistake.

Did you mean

$lim_(x rarr -3) (x^2+x-6)/(x+3)$

that in fact gives $0/0$ ?

Calculate the division of $x^2+x-6$ by $x+3$ which gives $x-2$ .

$lim_(x rarr -3) (x^2+x-6)/(x+3)=lim_(x rarr -3)((x-2)cancel((x+3)))/cancel(x+3)=lim_(x rarr -3) (x-2)=-3-2=-5$

How do you find the limit of f(x) = (x^2 + x - 6) / (x + 3) f(x)=x2+x−6x+3f(x) = (x^2 + x - 6) / (x + 3) as x approaches 0?