Limits · Lesson 1 of 5 · ~6 min

What a limit measures

Predict

$f(x) = \dfrac{\sin x}{x}$ is undefined at $x = 0$ — substitution gives $\frac{0}{0}$. As $x$ gets closer and closer to $0$, what do the outputs do?

Hold that thought. Drag offset h in the instrument and watch what the two markers do as $h$ shrinks.

Try

Try itDrag offset h from $2.00$ down toward $0.05$. The two markers at $x = -h$ and $x = +h$ ride the curve toward the open square; watch the f(h) and |f(h)−1| readouts as they go.

Notice

At $h = 2.00$ the readout shows $f(h) = 0.4546$. Slide down: $f(0.50) = 0.9589$, $f(0.10) = 0.9983$, $f(0.05) = 0.9996$, and the gap $|f(h)-1|$ collapses to $0.00042$. Meanwhile $f(-h)$ matches $f(h)$ at every stop. The outputs crowd toward $1$ from both sides — while the curve never touches the open square at $(0, 1)$.

Explain

You just measured a limit. The measurement never used a value at $x = 0$ — there is none. It used only nearby outputs, and they crowd toward one number, $1$. That is what a limit records: the trend of the nearby outputs, not the value at the point. $f(0)$ can be missing and the trend is still there.

So the prediction had a definite answer even though substitution failed. In symbols:

\lim_{x \to 0} \frac{\sin x}{x} = 1

Apply

Checkpoint — prove it to yourself

The instrument showed $f(0.05) = 0.9996$ while $f(0)$ does not exist. What does $\lim_{x \to 0} \frac{\sin x}{x} = 1$ claim?

Using the trend you measured, evaluate $\lim_{x \to 0}\ 2 \cdot \frac{\sin x}{x}$.

Transfer: $g(x) = \frac{x^2 - 9}{x - 3}$ is undefined at $x = 3$. Nearby, $g(2.9) = 5.9$ and $g(3.1) = 6.1$. What is $\lim_{x \to 3} g(x)$?

Key idea

A limit reports where the nearby outputs are heading, never what the function equals at the point. $f(a)$ and $\lim_{x \to a} f(x)$ answer different questions — and only the second needs to make sense.

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