How do you find the limit of $\frac{3 x^{2} + 6 x}{x^{2} - 4}$ $(3x^2+6x)/(x^2-4)$ as x approaches 2?

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Answer 1 · 2018-04-09T17:13:02

Please see below !

$3 x^{2} + 6 x = 3 x (x + 2)$ $3x^2+6x=3x(x+2)$

$x^{2} - 4 = (x - 2) (x + 2)$ $x^2-4=(x-2)(x+2)$

$(3xcancel((x+2)))/((x-2)cancel((x+2)))$

$(3x)/(x-2)$

I think you have Wrote the question wrongly and it approaches to $(-2)$ instead

Or if it as you said approaches to 2 So, the answer is D.N.E (doesn't exist

look at the graph

the limit for the 2 from the left is $-oo$ and from right $oo$

SO IT DOEN't exist.

graph{(3x^2+x6)/(x^2-4) [-8.38, 11.62, -9.28, 0.72]} )

BUT if it is approaches to -2 it gice you a value ( $(3x)/(x-2)$ ) BLug in here,

How do you find the limit of (3x^2+6x)/(x^2-4)3x2+6xx2−4(3x^2+6x)/(x^2-4) as x approaches 2?