Question

When would lim(x->1)(f(x))=0 but lim(x->1)(f(x)/g(x))=8?

Answer 1

we don't know what you are talking about.

Answer 2

so by the general limit: lim(x->1)(f(x)/g(x))=8 , we know that f(x) has to be 8 times g(x).
alternately, g(x) = 1/8*f(x)

from the first limit, lim(x->1)(f(x))=0
we know f(x) has to approach zero at the point x=1

We can really pick lots of functions here. I'm going to keep it fairly simple and use the example: f(x) = x-1

We know this function = 0 at x=1 so the first limit is satisfied.

We also already established with the second limit that g(x) should be 1/8 times f(x) at the point x=1.

Since we have no real criteria for what the functions should be, lets keep it simple and say that g(x) = 1/8*f(x) at all points. Therefore, g(x)=1/8*(x-1)

Now, even though g(x) is zero at x=1 and there is a singularity at x=1 of f(x)/g(x), the limit from either side still approaches 8, therefore the general limit still approaches 8