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?

[User1php]

(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?

 

[Dr.Peterson]

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?

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.)

But there is no such rule. If there is no limit, then there is no limit. You could mean that sometimes, when f(x) appears not to have a limit (because it is an indeterminate form), there may be a way around that; but that is nothing like what you wrote.

Maybe you need to give a specific example of what you are looking for.

 

[User1php]

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)

 

[Bliman]

(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?

Maybe you mean L'Hopital's rule https://en.wikipedia.org/wiki/L'Hôpital's_rule

 

[User1php]

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)
(first message i sent which is a copy of this is awaiting moderator approval -_-)

 

[Steven G]

Here is one.

Let f(x) = g(x) = [math] \dfrac {|x-5|}{x-5}[/math]

 

[User1php]

Here is one.

Let f(x) = g(x) = [math] \dfrac {|x-5|}{x-5}[/math]

(i think you misunderstood the question)

 

[Dr.Peterson]

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)

(i think you misunderstood the question)

The trouble is, you're a little vague about the question yourself.

What you said here doesn't really match the original question; but it's true that two functions may not have limits, but their product does. (This is not always true.)

Jomo gave an example of that :

Here is one.

Let f(x) = g(x) = [math] \dfrac {|x-5|}{x-5}[/math]

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.

Most likely, what you asked is really not what you want to ask, but you're having trouble expressing it. Maybe you need to try again.

 

[Steven G]

(i think you misunderstood the question)

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.