Question

How would you "remove the discontinuity" of f? In other words, how would you define
f(5) in order to make f continuous at 5?

f(x) = [(x^2 -2x -15)/x-5]

Answer 1

plug the hole at x = 5

Answer 2

how would you do that, that is what i am confused on

Answer 3

you first need to find the value of the limit at x=5
so\[\lim_{x \rightarrow 5} {{x^2-2x-15}\over{x-5}}\]
\[\lim_{x \rightarrow 5} {(x-5)(x+3)\over(x-5)}\]
after cancelling, the limit is 8
therefore you need to describe you function as
\[f(x)=|^{x^2-2x-15 , x \neq 5}_{8, x=5} \]

Answer 4

ah i see, i didnt realize i just had to find the limit..i was over thinking, thank you for your help!