This makes no sense. I'm guessing that you mean, there is still a way to find the limit. (There is no one value of x in this expression to be found; it is just a variable, approaching some specified number.)What was the rule in limits that if lim f(x) x->anynumberhere doesn't exist then there is still a way to find x?
Maybe you mean L'Hopital's rule https://en.wikipedia.org/wiki/L'Hôpital's_rule(i forgot and couldn't find it again in khanacademy) What was the rule in limits that if lim f(x) x->anynumberhere doesn't exist then there is still a way to find x?
(i think you misunderstood the question)Here is one.
Let f(x) = g(x) = [math] \dfrac {|x-5|}{x-5}[/math]
Like this:
lim f(x) x->5 and lim g(x) x->5 dont exist but
lim f(x)*g(x) x->5 exists. (i think it was something like that)
The trouble is, you're a little vague about the question yourself.(i think you misunderstood the question)
Here, f and g do not have limits as x approaches 5 (try it!); but their product, [MATH]\dfrac {|x-5|^2}{(x-5)^2}[/MATH], is equal to 1 everywhere except at 5, so it has a limit.Here is one.
Let f(x) = g(x) = [math] \dfrac {|x-5|}{x-5}[/math]
I think that you should look at the the three limits more closely and you'll see that my example matches exactly what you asked for.(i think you misunderstood the question)