I know how to check continuity of a function at a given point. I also know that a function is said to be continuous if it is continuous at every point in its domain. When I am asked if a function is continuous, for example $f(x,y)=\frac{2}{5x^3y}$, do I first find the domain and see if there are any points of discontinuities in the domain? In this case, domain of $f(x,y)=\{(x,y)\in \mathbb{R}^2:x\ne 0 \text{ or } y \ne 0\}$. Will I be correct to say that f(x,y) is continuous since the points of discontinuities are not in the domain of f(x,y)?
continuity of a function without a specified point
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calculus
2 Answers
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To say that "$f(x,y)$ is continuous since the points of discontinuities are not in the domain of $f(x,y)$" is not perfectly rigorous (since the function simply does not exist outside its domain, so you can't even say that there are discontinuities in that nowhere), but it captures the correct intuition.
You only care about what happens in the domain of the function; so, for example, if you have a function that is continuous somewhere and discontinuous elsewhere, and you remove from the domain all points of discontinuity, you are left with a continuous function.
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no, there are no points of discontinuities at all here!
to be continuous or not, first the function has to be defined at a point! (if not defined, you can't say it is continuous or discontinuous -- see the definition of continuity)
in this example, you can say that $f$ is continuous since it is a quotient of two continuous functions
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0given $f(x)=\frac{x^2-1}{x-1}$, I think that even though f(x) is not defined at x=1, I can still say f(x) is not continuous at x=1. For this, I reffer you to Tomas calculs 11th edition page 127, read example 5 please. – 2017-02-27