Locating discontinuities in functions My professor does not bother to explain how to do it, but bothers to arrange a quiz... so here is my question
How to locate a point of didcontinuiity
Find all points of discontinuity:
$f(x) = (x^2-3)/(x^2+2x-8)$
I would really appreciate it if you could also explain how you came up with thr answer briefly, because i am finding the chapter continuity extremely difficult.
 A: All polynomials are continuous in $\mathbb R$.
All rational fractions (ratios of polynomials) are continuous in their domain (the whole of $\mathbb R$, except the zeroes of the denominator).
This is because the sum, difference, product and quotient (except where $0$) of two continuous functions are continuous functions, and $x$ and constants are continuous functions.
A: $f(x)$ has no points of discontinuity, but there are points on the real line $\Bbb R$ where $f$ is undefined.  But being undefined is not the same as being discontinuous.

Fun fact:  All elementary functions are continuous on their domains.
(Note that this excludes piecewise functions, but I don't think piecewise functions technically count as elementary functions anyway.. Not entirely sure on that.)
$f(x) = \dfrac{x^2-3}{x^2+2x-8}$ is an elementary function.
Therefore $f(x)$ is continuous on its domain, i.e., $f(x)$ is continuous everywhere it's defined.
$f(x)$ is defined everywhere except where $x^2 + 2x - 8 = 0$.
$x^2 + 2x - 8 = 0$ exactly when $x = 2$ or $x=-4$.
So $f(x)$ is defined on $(-\infty, -4) \cup (-4, 2) \cup (2, +\infty)$.
Therefore $f(x)$ is continuous on $(-\infty, -4) \cup (-4, 2) \cup (2,+\infty)$.
(Some people might say that last line is incorrect and will instead insist that we say "$f(x)$ is continuous on $(-\infty, -4)$, continuous on $(-4,2)$, and continuous on $(2, +\infty)$."  Just know what your instructor wants.)
It is not correct to say $f(x)$ is discontinuous at $x=-4$ and $x=2$.  It's incorrect to say that because $f(x)$ is not even defined there.  For a function to be discontinuous at a point, the function must be defined at that point.
A: The answers given so far perhaps should be directed, not to the OP, but to OP's instructor or to the author of OP's textbook
It is a fact that most elementary calculus textbooks define "continuity at a point" in such a way that "point" means a point in $\mathbb{R}$. The standard definition in such standard texts is
The statement that the function $f$ is continuous at the point $a\in\mathbb{R}$ means that


*

*$a$ lies in the domain of $f$

*$\lim_{x\to a}f(x)$ exists

*$\lim_{x\to a}f(x)=f(a)$


Given this definition of continuity at a point (of $\mathbb{R}$) the function
$$ f(x) = \dfrac{x^2-3}{(x+4)(x-2)} $$
clearly fails both the first and the second conditions of the definition at $a=-4$ and $a=2$.
