VALUE OF THE FUNCTION
The most direct way to probe a function at a point is to simply evaluate it there — substitute the input and read off the output. If $a$ is the point of interest, the value of the function is $f(a)$. This is the function answering a single, specific question: “what is $y$ when $x$ is exactly this?”
$f(x) = x^2 – 1$
$$f(3) = 3^2 – 1 = 8$$
$f(x) = \sqrt{x}$
$$f(9) = 3$$
$f(x) = \dfrac{1}{x}$
$$f(4) = \frac{1}{4}, \qquad f(0) \text{ does not exist}$$
ZEROS — WHERE THE FUNCTION CROSSES THE X-AXIS
A zero of a function is a point where the probe finds for which $x$, the function becomes zero or $f(x) = 0$. Geometrically, this is where the curve meets the x-axis — the boundary between the function being positive and being negative. Finding zeros is really finding the solutions of the equation $f(x) = 0$, and a function can have no zeros, one, or several, depending on its form.
$f(x) = x^2 – 1$
$$f(x) = 0 \implies x = 1, \ x = -1$$
$f(x) = x – 5$
$$f(x) = 0 \implies x = 5$$
$f(x) = e^x$
$$e^x = 0 \implies \text{no solution}$$
$f(x) = x^2 + 1$
$$x^2 + 1 = 0 \implies \text{no real solution}$$
LIMIT — APPROACHING FROM THE LEFT AND THE RIGHT
Sometimes the question isn’t “what is the function exactly at this point” but “what is it heading toward as we get close to this point.” This is the idea of a limit. Since a point on the number line can be approached from two directions, the probe splits into a left-hand limit and a right-hand limit — what the function does just before $a$, and just after $a$. The limit at $a$ exists only when both of these approaches agree.
$f(x) = x^2 – 1$
$$\lim_{x \to 2^-} f(x) = 3, \qquad \lim_{x \to 2^+} f(x) = 3, \qquad \lim_{x \to 2} f(x) = 3$$
$f(x) = \sin x$
$$\lim_{x \to 0^-} f(x) = 0, \qquad \lim_{x \to 0^+} f(x) = 0, \qquad \lim_{x \to 0} f(x) = 0$$
$$f(x) = \begin{cases} 1 & x < 0 \\ 2 & x \geq 0 \end{cases}$$
$$\lim_{x \to 0^-} f(x) = 1, \qquad \lim_{x \to 0^+} f(x) = 2, \qquad \lim_{x \to 0} f(x) \text{ does not exist}$$
WHERE VALUE AND LIMIT MIGHT FAIL
Both probes — value and limit — can fail, and they fail for different reasons.
The value $f(a)$ can fail to exist simply because $a$ is not in the domain at all, or because the function has a gap or a break exactly there.
The limit can fail to exist for a different reason: the left-hand and right-hand approaches may not agree, or the function may behave too erratically near $a$ to settle toward anything at all.
$f(x) = \dfrac{1}{x-2}$
$$f(2) \text{ does not exist}$$
$$f(x) = \begin{cases} 1, & x < 0 \\ 2, & x \geq 0 \end{cases}$$
$$\lim_{x \to 0^-} f(x) = 1, \qquad \lim_{x \to 0^+} f(x) = 2, \qquad \lim_{x \to 0} f(x) \text{ does not exist}$$
$f(x) = \sin\left(\dfrac{1}{x}\right)$
$$\lim_{x \to 0} f(x) \text{ does not exist (oscillates indefinitely)}$$
WHAT’S LOST WHEN EITHER (OR BOTH) IS MISSING
When the value is undefined, the function simply has nothing to say at that point — there’s no $y$ to report.
When the limit fails to exist, the function has no consistent behaviour to understand near that point, even if a value happens to exist there.
Losing one of these is a localized problem; losing both means the point offers no information whatsoever, either about itself or its surroundings.
$f(x) = \dfrac{1}{x-2}$
$$f(2) \text{ does not exist}, \qquad \lim_{x \to 2^-} f(x) = -\infty, \qquad \lim_{x \to 2^+} f(x) = +\infty$$
$$f(x) = \begin{cases} x+1, & x \neq 3 \\ 0, & x = 3 \end{cases}$$
$$\lim_{x \to 3} f(x) = 4, \qquad f(3) = 0$$
$f(x) = \sqrt{x}$
$$\lim_{x \to 0^-} f(x) \text{ does not exist (undefined to the left)}, \qquad \lim_{x \to 0^+} f(x) = 0, \qquad f(0) = 0$$
CONTINUITY
Continuity at a point is the statement that all three probes agree: the left-hand limit, the right-hand limit, and the value of the function at that point are all the same number.
When this agreement holds at every point of an interval, the function is continuous there — and continuity is precisely the absence of every failure described above.
$f(x) = x^2 – 1$
$$\lim_{x \to 2^-} f(x) = 3, \qquad \lim_{x \to 2^+} f(x) = 3, \qquad f(2) = 3$$
$$f(x) = \begin{cases} x+1, & x \neq 3 \\ 0, & x = 3 \end{cases}$$
$$\lim_{x \to 3} f(x) = 4, \qquad f(3) = 0 \qquad \implies \text{not continuous at } x = 3$$
$$f(x) = \begin{cases} 1, & x < 0 \\ 2, & x \geq 0 \end{cases}$$
$$\lim_{x \to 0} f(x) \text{ does not exist} \qquad \implies \text{not continuous at } x = 0$$
This tool will help to visualize these ideas