TITLE: Why is $y=1/x$ a continuous function but not $y=(1/x)^2$?
QUESTION [12 upvotes]: I was reading Thomas' calculus which said that $y=1/x$ is a continuous function because it was continuous at every point of its domain(it not being defined at $x=0$), but then goes on to show that $1/x^2$ is discontinuous saying it is an infinity discontinuity as $x$ approaches $0$. Why doesn't the same logic apply to $y=1/x$?

REPLY [5 votes]: Short answer: The same logic does apply for $y = \frac{1}{x}$.
Long answer: A continuous function is defined to have no discontinuities within its domain. Therefore, $y = \frac{1}{x}$ is a continuous function because $x = 0$ is not part of its domain. However, there is an infinite discontinuity at $x = 0$, but because $x = 0$ is not part of the function's domain, $y = \frac{1}{x}$ is still considered to be a continuous function. Using the same logic, $y = \frac{1}{x^2}$ is also a continuous function, just with a discontinuity at $x = 0$, which is not part of its domain.