Answer:
2. [tex]f(x)=\frac{x^2}{x+5}[/tex]
Step-by-step explanation:
When you replace x=5 into those functions you obtain:
[tex]1. f(x)=\sqrt{x^2-36} = \sqrt{5^2-36} = \sqrt{-11} \\2. f(x)=\frac{x^2}{x+5}=\frac{5^2}{5+5}=\frac{25}{10}[/tex]
We can see that the function number 1, when the x=5 is replaced, falls into the imaginary numbers or complex domain, so the function is not continuous in the real domain.
On the other hand, the second function falls into the real domain whe x=5 is replaced, so the function is continuous in the real domain!
I hope you understand!
