Check the one-sided limits:
[tex]\displaystyle \lim_{x\to5^-}f(x) = \lim_{x\to5}kx = 5k[/tex]
[tex]\displaystyle \lim_{x\to5^+}f(x) = \lim_{x\to5}8x^2 = 200[/tex]
If f(x) is to be continuous at x = 5, then these two limits should have the same value, which means
5k = 200
k = 200/5
k = 40
