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

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Answer 1 · 2018-04-09T17:13:02

Please see below !

#3x^2+6x=3x(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,