Learning how to determine where a function is continuous requires understanding the relationship between a function’s value and its limit at a point. A function is continuous at a point when those two quantities agree, meaning there is no break, jump, hole, or sudden change in its behavior there. This concept is essential in calculus, graph analysis, physics, economics, engineering, and computer science.
Short version: it depends. Long version — keep reading.
The Definition of Continuity
A function (f) is continuous at (x=a) if all three conditions are satisfied:
- (f(a)) is defined.
- (\lim_{x\to a}f(x)) exists.
- The limit equals the function value:
[ \lim_{x\to a}f(x)=f(a) ]
The first condition is often overlooked, but it is just as important as the other two. Even if a limit exists at a point, the function is not continuous there if its value is undefined.
Here's one way to look at it: consider
[ f(x)=\frac{x^2-4}{x-2} ]
The expression can be simplified:
[ f(x)=\frac{(x-2)(x+2)}{x-2}=x+2 ]
That said, this simplification is valid only when (x\neq2). Because of this,
[ \lim_{x\to2}f(x)=4, ]
but (f(2)) is undefined. So the function is continuous for every real number except (x=2). At (x=2), the graph has a removable discontinuity, commonly called a hole And that's really what it comes down to..
Why the Three Conditions Matter
The three-part definition helps distinguish several different situations.
The Function Value Is Undefined
If (f(a)) does not exist, the function cannot be continuous at (a). This often happens when the denominator of a rational function is zero.
To give you an idea,
[ g(x)=\frac{1}{x-4} ]
is not defined at (x=4). Its behavior near (4) also shows why:
[ \lim_{x\to4^-}g(x)=-\infty, \qquad \lim_{x\to4^+}g(x)=+\infty ]
This is an infinite discontinuity, and the graph has a vertical asymptote at (x=4) It's one of those things that adds up. Less friction, more output..
The Limit Does Not Exist
Even when (f(a)) is defined, the function may be discontinuous because the two-sided limit fails to exist. This commonly occurs when the left-hand and right-hand limits are different Which is the point..
As an example, let
[ h(x)= \begin{cases} 0, & x<1,\ 1, & x\geq1. \end{cases} ]
At (x=1),
[ \lim_{x\to1^-}h(x)=0 ]
but
[ \lim_{x\to1^+}h(x)=1. ]
Because the one-sided limits do not agree, the two-sided limit does not exist. The function has a jump discontinuity at (x=1) Not complicated — just consistent..
The Limit Exists but Does Not Equal the Function Value
Suppose
[ f(x)= \begin{cases} x^2, & x\neq3,\ 100, & x=3. \end{cases} ]
The limit at (x=3) is
[ \lim_{x\to3}f(x)=3^2=9. ]
On the flip side, (f(3)=100). Since (9\neq100), the function is discontinuous at (x=3). This is another removable discontinuity because changing the value of (f(3)) to (9) would make the function continuous there Most people skip this — try not to..
