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

Question

1776 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-05T13:38:01

0

#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#