Answer:
Yes, its continuous
Step-by-step explanation:
We use the formula:
x^2-y^2=(x-y)(x+y),
And we know that 16=4^2, so we have:
[tex]\frac{x^2-16}{(x+4)(x-5)}=\frac{(x-4)(x+4)}{(x+4)(x-5)}=\frac{x-4}{x-5}[/tex]
So for x=-4 we have -8/-9,i.e, it is 8/9, so it is continuous.
I dont know what is x=1, because for x=1 the function has value 3/4.
But function is not continuous in x=5 becaus for that x we will get 1/0, and that is not definite.
:)
