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I used x^2 + 2(x)(h) + h^2 + x +h -3 -x^2 -x +3 = 2x + x ??

basically factor out the h and H(h+2x+1) divided by the h on denominator and = h + 2x + 1?

Yes.

And you got:
\[\lim_{h \rightarrow 0} (h+2x+1)\]Can you evaluate the limit?

Would be 2x + 1 ??? Because h is going to 0

Indeed it is!

Holy Cow! It came together finally!! Thank you very much for your help! I would have been stuck!!!

You're welcome :)

Have a great day!!

You too! :)