How do you evaluate the limit of #lim (h^2+4h+4)/(h+2)# as #h->-2#?

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Answer 1 · 2016-12-05T13:38:01

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#Lim_(hrarr-2)(h^2+4h+4)/(h+2)#

#Lim_(hrarr-2)((h+2)cancel((h+2)))/cancel((h+2)#

#Lim_(hrarr-2)(h+2)=0#