What is the limit of #(2x^2-18) / (x+3)# as x approaches -3?

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Answer 1 · 2016-06-18T15:49:10

4

when seeing this type of question #0/0#
you need to use L Hospital LAW
#lim_(x->-3)(2x^2-18)/(x+3)#
=#lim_(x->-3)(4x)/x#
=#4#

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May I suggest the following solution ...Tony B

Look at limit ex Maple =-12
Tony B

Limit ex EfOfEx
Tony B

If you apply polynomial division you end up with #2x-6#

Or if you factor you have: #(2(x^2-3^2))/(x+3)#

#(2(x-3)cancel((x+3)))/cancel((x+3)) = 2x-6#

So at #x=-3" we have "2(-3)-6 = -12#