Answer:
(i) and (iii)
Step-by-step explanation:
for i)
f(x) = [tex]\frac{x^{2}-4}{x+2}[/tex]
=[tex]\frac{(x+2)(x-2)}{x+2}[/tex]
=(x-2)
As x =-2
f(x)=-2-2
f(x)=-4
so it is a continuous function
for ii)
from the result of (i) we can deduce the result of second and we see that it is not equal to zero so it is not a continuous function
iii)
f(x)=[tex]\frac{x^{2}+4x+4}{x+2}[/tex]
f(x)=[tex]\frac{(x+2)^{2}}{x+2}[/tex]
f(x)=[tex]\frac{(x+2)(x+2)}{x+2}[/tex]
f(x)=x+2
as x= -2
f(x)=-2+2
f(x)=0
so it is a continuous function
iv) From the result of iii we can deduct that
f(x)=0 because it have the same equation so it is not a continuous function
