For #x<-2, f(x)=(x^2+6x+8)/(x+2)=((x+4)(x+2))/(x+2)#, and, since,
#x!=-2#, we can cancel #(x+2)# to get,
#f(x)=((x+4)cancel(x+2))/cancel(x+2)=x+4, if, x<-2#
Now, as #xrarr -2-, x<-2, so, f(x)=x+4#
#:. lim_(xrarr-2-)f(x)=lim_(xrarr-2-)(x+4)=-2+4=2..........(1)#
Further, as #xrarr-2+, x>-2, so, f(x)=kx^2#
#:. lim_(#xrarr-2+) f(x)=lim_(#xrarr-2+)kx^2=4k...................(2)#
Also, #f(-2)=4k....................................(3)#
Because #f# is reqd. to be continuous at #x=-2#, by defn. of continuity, we must have,
#lim_(xrarr-2-)f(x)=f(-2)=lim_(xrarr-2+)f(x)#.
#(1),(2) and (3) rArr 2=4k rArr k=1/2#.
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