Question #e007a

1 Answer
Jim S
Dec 8, 2017

#x>##-2# then #lim_(xrarr-2)(3x+6)/|x+2|=3# , #x<##-2# then #lim_(xrarr-2)(3x+6)/|x+2|=-3#

Explanation:

  • If #x+2>0# #<=># #x>##-2# , #x->-2^+# then

#lim_(xrarr-2^+)(3x+6)/(|x+2|)# = #lim_(xrarr-2^+)(3x+6)/(x+2)# #=#
#=# #lim_(xrarr-2^+)3(x+2)/(x+2)# = #3lim_(xrarr-2^+)cancel(x+2)/cancel(x+2)# #<=># #3lim_(xrarr-2^+)1# #=3#

  • If #x+2<0# #<=># #x<-2# , #x->-2^-#

#lim_(xrarr-2^-)(3x+6)/(|x+2|)# = #lim_(xrarr-2^-)(3x+6)/-(x+2)# #=#
#=# #-3lim_(xrarr-2^-)cancel(x+2)/cancel(x+2)# #=-3*1=-3#

Related questions
Impact of this question
738 views around the world
You can reuse this answer
Creative Commons License