Is a function with a removable discontinuity considered continuous? Take for example $\frac{x^2-4}{x-2}$. It reduces to $x+2$.
Asked
Active
Viewed 147 times
2 Answers
2
If the discontinuity is not fixed, the function will remain discontinuous, because we don't have that
$\displaystyle\lim_{x\mathop\rightarrow a}f(x)=f(a)$, this because $f(a)$ is undefined.
Daniel Charry
- 345
-
If you simplify the function before calculating f(a) is it not equal to the limit of f(x)? Or is x+2 considered an entirely different function? – user135247 Jul 20 '14 at 11:39
0
It is continuous if you consider $f$ defined "away" from the problem, that is, $\mathrm{Dom}(f) = \Bbb R \setminus (2 - r, 2 + r)$, for some $r > 0$. However, if you consider $\mathrm{Dom}(f) = \Bbb R \setminus \{2\}$, then it is not continuous. To make it continuous on this second situation, you should define the value of the function on the problematic point, that in your case is $x=2$. But we can't define it just like this, happy-go-lucky. We must have also that $$\lim_{x \to 2} f(x) = f(2)$$ If we have: $$f(x) = \left\{\begin{array}{cc} \frac{x^2 - 4}{x - 2} & \mbox{if } x \neq 2 \\ 4 & \mbox{if } x = 2\end{array}\right.$$ then $f$ is continuous.
Ivo Terek
- 77,665