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

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

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Answer 1 · 2015-04-23T18:23:02

I think that it diverges giving you:
#lim_(x->2^-)(4x^2)/(x-2)=-oo#
#lim_(x->2^+)(4x^2)/(x-2)=+oo#
By substituting values near #2# from the felt and right I got:
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How do you find the limit of #(4x^2)/(x-2)# as x approaches 2?

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