Limits Revisited

By the end you should be able to: State what a limit means informally and formally, use one-sided limits to decide whether a limit exists, and explain why the value of f(a) is irrelevant to the limit at a.

You met limits in Plus-Two. This is the same idea stated precisely enough to build the rest of calculus on.

The idea in one sentence

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

means: as $x$ gets close to $a$ , $f(x)$ gets close to $L$ — and we can make $f(x)$ as close to $L$ as we like by taking $x$ close enough to $a$ .

The part students consistently get wrong

The limit at $x = a$ says nothing about $f(a)$ .

It is about the values of $f$ at points near $a$ , deliberately excluding $a$ itself. Three separate situations, all with the same limit:

Situation	$f(2)$	$\lim_{x\to 2} f(x)$
$f(x) = x + 1$	3	3
$f(x) = \dfrac{x^2 - 1}{x - 1}$ at $x=2$	3	3
$g(x) = x+1$ for $x \neq 2$ , $g(2) = 100$	100	3

In the third row the function is defined at 2, and the limit still ignores it. The limit is what you would expect the value to be from the surrounding behaviour, whether or not the function obliges.

The formal definition

$\lim_{x \to a} f(x) = L$ means: for every $\varepsilon > 0$ there is a $\delta > 0$ such that

0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon

Read it as a challenge and a response. Someone challenges you with a tolerance $\varepsilon$ — "get within this much of $L$ ". You must respond with a $\delta$ — "stay within this much of $a$ and you will". If you can always answer, the limit exists.

Note $0 < |x - a|$ . That strict inequality is what excludes $x = a$ itself.

One-sided limits, and the existence test

\lim_{x \to a^-} f(x) \quad \text{approach from the left} \qquad \lim_{x \to a^+} f(x) \quad \text{approach from the right}

The test: the two-sided limit exists if and only if both one-sided limits exist and are equal.

\lim_{x\to a} f(x) = L \iff \lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = L

The standard counterexample

f(x) = \frac{|x|}{x}

For $x > 0$ this is $x/x = 1$ . For $x < 0$ it is $-x/x = -1$ .

\lim_{x\to 0^-} \frac{|x|}{x} = -1, \qquad \lim_{x\to 0^+} \frac{|x|}{x} = 1

The one-sided limits disagree, so $\lim_{x\to 0} |x|/x$ does not exist. The graph jumps from $-1$ to $1$ with nothing bridging the gap.

The same reasoning applies to the unit step function, and to any function defined piecewise where the pieces do not meet.

Why this matters for what follows

The derivative is defined as a limit. Continuity is defined using a limit. If the limit machinery is shaky, everything built on it is too — which is why this apparently basic topic is the first thing in the syllabus.