 # If f(x)=1/(x−2)(2x−5) and g(x)=1/x^2 , we define f(g(x)) over the domain in which its defined , is it correct to say f(g(x)) is discontinuous at +−sqrt(2/5) ,0 and +-1/sqrt 2 ? 0 because the input is going into g(x) hence as its not in domain of g(x) so it can be said its discontinuous at that point . and similarily other four points as its not in domain of f(g(x)) ? But if suppose we say h(x)=f(g(x)) will we say h(x) is discontinuous at those points (maybe different ones) where x is not defined after simplyfing the whole f(g(x)) to a rational function ? janine83fz 2022-08-12 Answered
If $f\left(x\right)=\frac{1}{\left(x-2\right)\left(2x-5\right)}$ and $g\left(x\right)=\frac{1}{{x}^{2}}$ , we define f(g(x)) over the domain in which its defined , is it correct to say f(g(x)) is discontinuous at $+-\surd \left(2/5\right)$ ,0 and +-$1/\surd 2$ ? 0 because the input is going into g(x) hence as its not in domain of g(x) so it can be said its discontinuous at that point . and similarily other four points as its not in domain of f(g(x)) ? But if suppose we say h(x)=f(g(x)) will we say h(x) is discontinuous at those points (maybe different ones) where x is not defined after simplyfing the whole f(g(x)) to a rational function ?
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it tangouwn
The definition of "h is continuous at a" is
$\underset{x\to a}{lim}h\left(x\right)=h\left(a\right)$
For this to be true, three things have to be satisfied:

1. h(x) must converge to some limit L as $x\to a$.
2. h(a) must be defined. That is, a is in the domain of h.
3. The values h(a) and L must be the same.

Since none of $±\sqrt{\frac{2}{5}},±\frac{1}{\sqrt{2}},0$ are in the domain of h, h is not continuous at any of those points.
But does that mean the same thing as saying "h is discontinuous" at those points? I would argue no. If we take "h is discontinuous at a" to be exactly the negation of "h is continuous at a", then we can equally well say "h is discontinuous at Sally", because Sally is also not in the domain of h.
But that is silly. h is defined on a subset of the real numbers, so most things are not going to be in the domain of h. Saying "h is discontinuous at a" should mean something significant about the behavior of h, not be something trivial and useless. So to say "h is discontinuous at a" should mean h is defined at a, but either does not converge there, or else converges to some other value than h(a).
That is my opinion. But I am not the Lord High Arbiter of Mathematical Language. There are doubtless others who disagree. Some things are not quite important enough for a general consensus to develop.