LHD at x=3
x⟶3−lim(x−3kx+1−2k)
=x→3−limk(x−3x−3).x+1+21=4k
RHD at x=3
x→3+limx−3(mx+2)−2k
Since this limit exists if 3m+2−2k=0...(i) (00 form)
So,x→3+limx−3(mx+2)−(3m+2)=m
Hence m=4k…(ii)
From (i)&(\mathrm{ii})
m=\frac{2}{5} & k=\frac{8}{5} \Rightarrow k+m=2