Alternate way is to use L'Hopital's rule

f(x) = ax + b
f'(x) = a

lim (√x - 4)/f(x)
x→16

= lim d/dx(√x - 4) / f'(x)
x →16

= lim (1/2√x) / a
x → 16

= lim 1/(2a√x)
x → 16

= 1/(2a√16)

= 1/8a

So,

1/8a = 1/16
8a = 16
a = 2

Then,

lim (√x - 4)/f(x)
x → 16

= lim (√x - 4)/(2x + b)
x → 16



What we realise is that if we don't cancel common factors, the limit is actually indeterminate i.e 0/0 since numerator = √16 - 4 = 0

This tells us that the denominator has to be 0.
(If the denominator f(16) was non-zero, it would be impossible to get 1/16 as the limit. The limit would be 0/(non-zero) = 0)


So,

2(16) + b = 0
32 + b = 0
b = -32