Limits and Continuity | Moosa Academy

Limits and Continuity — Moosa Academy

Differentiation measures a rate of change; integration measures an area. Neither works where a curve tears or jumps. Continuity is the condition that rules that out, and limits are the tool used to test for it — which makes this lesson the doorway into calculus.

Concept The limit

\lim_{x \to c} f(x) = L

This says: as $x$ gets arbitrarily close to $c$ , the value of $f(x)$ settles on $L$ . Note what it does not say — it makes no claim about $f(c)$ itself. The function may not even be defined there.

Approaching from the right:

\displaystyle\lim_{x \to c^{+}} f(x)

Approaching from the left:

\displaystyle\lim_{x \to c^{-}} f(x)

The limit exists only if the two one-sided limits agree. If the curve heads for different values from each side, there is no single value to settle on.

Concept Continuity and its three conditions

A function is continuous if you can draw it without lifting your pen.

Stated precisely, $f$ is continuous at $c$ when all three hold:

$f(c)$ is defined.
$\displaystyle\lim_{x \to c} f(x)$ exists.
$\displaystyle\lim_{x \to c} f(x) = f(c)$ .

Fail any one and the function is discontinuous there.

Note Three kinds of discontinuity

Infinite — the curve runs off to ±∞, as

1/x

does at 0.

Removable — a single hole; the limit exists but the value is missing or wrong.

Jump — the one-sided limits differ, so the curve steps.

Only the removable kind can be repaired — redefine the function at that one point and the gap closes.

Example A removable discontinuity

Examine $f(x) = \dfrac{x^2 - 4}{x - 2}$ at $x = 2$ .

x = 2

the denominator is zero, so

f(2)

is undefined — condition 1 already fails.

Factorise:

\dfrac{(x-2)(x+2)}{x-2} = x + 2

for every

x \neq 2

\displaystyle\lim_{x \to 2} f(x) = 4

from both sides, so the limit does exist.

Defining

f(2) = 4

closes the hole.

⟹ removable discontinuity at x = 2

Example A jump discontinuity

Examine $f(x) = \begin{cases} x + 1 & x < 1 \\ x + 3 & x \geq 1 \end{cases}$ at $x = 1$ .

From the left:

\displaystyle\lim_{x \to 1^{-}} f(x) = 1 + 1 = 2

From the right:

\displaystyle\lim_{x \to 1^{+}} f(x) = 1 + 3 = 4

The one-sided limits differ, so no limit exists at

x = 1

⟹ jump discontinuity — the curve steps from 2 to 4

No redefinition can fix this: a single value cannot equal both 2 and 4.

Theorem The Intermediate Value Theorem

If $f$ is continuous on $[a,\ b]$ and $N$ lies between $f(a)$ and $f(b)$ , then $f(c) = N$ for some $c$ in the interval.

A continuous curve cannot skip values. To get from one height to another it must pass through everything in between.

f(a) = -5

and

f(b) = 3

, the function must cross zero somewhere between — which proves a root exists without finding it.

f(a) = 2

and

f(b) = 5

, the function must take the value 3 somewhere between.

Continuity is essential to the argument. A function that jumps can step straight over a value without ever taking it.

Summary

A limit describes where a function heads as x approaches a point, not its value there.
The limit exists only when the left and right one-sided limits agree.
Continuity at c needs f(c) defined, the limit to exist, and the two to be equal.
Discontinuities are infinite, removable or jump — only the removable kind can be repaired.