# Problem about continuity of rational function

#### alan6690

##### New member
For a rational function f(x)/g(x), I know that we can say it is continuous if both f(x) and g(x) is continuous while g(x) is also not 0.

So my question is: for the same rational function f(x)/g(x), can we say it is continuous if we know the limit of f(x) exists and equals to 0 at our point of interest while we know the denominator g(x) is not zero but it is dis-continuous? My doubt is: the rational function would be continuous and reach 0 anyway because the numerator is 0, so in this scenario would the continuity of the denominator affect the continuity of the rational function if the denominator is not zero.

Hope you understand my question. Thank you.

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#### Dr.Peterson

##### Elite Member
A rational function is a quotient of polynomials. You are evidently asking about functions defined as quotients in general.

#### alan6690

##### New member
Sorry for the wrong terminology. What I mean indeed is just a quotient of two different function.

I am not really sure how to find an example for my conjecture. This question just comes up in my head when I am doing some math question set.
What if I define:
f(x) = x
g(x) = 1+x^2 when x <> 0, g(x) = 3 when x=0
In this way, f(x)/g(x) still reach 0 from both ends when x=0, even though g(x) is not continuous at that point.

Then do we say f(x)/g(x) is continuous when x=0?

Thank you.

#### Dr.Peterson

##### Elite Member
In such a case, just write the definition of f/g in piecewise form and check the limit and value as usual.

But what you're observing here is just that where f(x)=0, there can be a removable discontinuity in g and f/g will still be continuous. Other kinds of discontinuity would matter.

#### Jomo

##### Elite Member
No no no, you can't say that a fraction is 0 if the numerator is 0!!!

A fraction equals 0 if the numerator is 0 AND the denominator is not 0.

0/undefined is not 0!

And 0!= 1!

#### Jomo

##### Elite Member
And 0!= 1!
OK, but what is 0!!! Is it (0!)!!, (0!!)!, ((0!)!)! or something else?

#### Subhotosh Khan

##### Super Moderator
Staff member
OK, but what is 0!!! Is it (0!)!!, (0!!)!, ((0!)!)! or something else?
0!!! = (0!)!! = 1!! = 1...... it is 1 all the way - like those elephants holding up the universe.....

#### Jomo

##### Elite Member
0!!! = (0!)!! = 1!! = 1...... it is 1 all the way - like those elephants holding up the universe.....
They are all 1? Amazing.

#### alan6690

##### New member
In such a case, just write the definition of f/g in piecewise form and check the limit and value as usual.

But what you're observing here is just that where f(x)=0, there can be a removable discontinuity in g and f/g will still be continuous. Other kinds of discontinuity would matter.
Sorry for the late reply. And thank you! I think "removable discontinuity" is the thing I should look up for.

However, I am still a little bit confused about what you said by writing the definition of f/g in piecewise form.
In such form, should I write:
f/g = 0 when x=0
f/g = (x)/(1+x^2) when x <>0
If so, both pieces would just return the same value, which is 0, when x=0. Then it seems that the first piece is unnecessary.

Please kindly correct me for any mistake. Thank you.

#### alan6690

##### New member
No no no, you can't say that a fraction is 0 if the numerator is 0!!!

A fraction equals 0 if the numerator is 0 AND the denominator is not 0.

0/undefined is not 0!
I don't think the denominator g(x) would be 0 at any x as I define g(x) = 1+x^2

#### HallsofIvy

##### Elite Member
For a rational function f(x)/g(x), I know that we can say it is continuous if both f(x) and g(x) is continuous while g(x) is also not 0.

So my question is: for the same rational function f(x)/g(x), can we say it is continuous if we know the limit of f(x) exists and equals to 0 at our point of interest while we know the denominator g(x) is not zero but it is dis-continuous?
A rational function is, by definition, a polynomial over a polynomial. A polynomial is NEVER discontinuous so that cannot happen,

My doubt is: the rational function would be continuous and reach 0 anyway because the numerator is 0, so in this scenario would the continuity of the denominator affect the continuity of the rational function if the denominator is not zero.

Hope you understand my question. Thank you.

#### Dr.Peterson

##### Elite Member
However, I am still a little bit confused about what you said by writing the definition of f/g in piecewise form.
In such form, should I write:
f/g = 0 when x=0
f/g = (x)/(1+x^2) when x <>0
If so, both pieces would just return the same value, which is 0, when x=0. Then it seems that the first piece is unnecessary.
Initially, you need the piecewise form because g is defined that way. But then you can observe that the x=0 case can be absorbed into the other, as you say.

#### alan6690

##### New member
Initially, you need the piecewise form because g is defined that way. But then you can observe that the x=0 case can be absorbed into the other, as you say.
Yes that's what I mean.

So, in this scenario where f(x) is continuous, g(x) is not continuous. Can we say f(x)/g(x) is continuous?

#### Dr.Peterson

##### Elite Member
So, in this scenario where f(x) is continuous, g(x) is not continuous. Can we say f(x)/g(x) is continuous?
Yes.