0

If $f(x)=\frac{1}{(x-2)(2x-5)}$ and $g(x)=\frac{1}{x^2}$ , we define $f(g(x))$ over the domain in which its defined , is it correct to say $f(g(x))$ is discontinuous at $+-√(2/5)$ ,$0$ and +-$1/√2$ ? $0$ 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 ?

1 Answers1

1

The definition of "$h$ is continuous at $a$" is $$\lim_{x \to a} h(x) = h(a)$$

For this to be true, three things have to be satisfied:

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

Since none of $\pm\sqrt{\frac 25}, \pm\frac 1{\sqrt 2}, 0$ 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.

Paul Sinclair
  • 43,643
  • What about the f(g(x)) discontinuity Sir ? Will it include 0 or not and reason too ? I was looking for difference in the number of discontinuous points in h(x) which is =f(g(x)) and f(g(x)) , if we define like h(x) will it include zero or not ? – ProblemDestroyer Mar 23 '22 at 00:01
  • By definition $h(x) = f(g(x))$ so the answer for $h$ is the answer for $f\circ g$. But you are right that I forgot to include $0$ as point where $h$ is not defined, and so is not continuous. Since $g(0)$ is not defined, neither can $f(g(0))$ be. – Paul Sinclair Mar 23 '22 at 00:04