Write $\displaystyle\lim_{x\to a} f(x) = L$ to mean: as $x$ gets closer and closer to $a$ (from both sides), $f(x)$ gets closer and closer to the single number $L$ . The phrase “from both sides” is the whole game — the left approach and the right approach must agree.

The value at the point is irrelevant

This is the idea students fight hardest. $\displaystyle\lim_{x\to a}f(x)$ does not care what $f(a)$ is — or whether $f(a)$ even exists. It only cares where $f$ is headed. The three graphs below all have $\lim_{x\to 2}f(x)=3$ , even though $f(2)$ is different (or missing) in each.

[Diagram — see the figure in the print workbook.]

When a (two-sided) limit fails to exist

A limit does not exist (DNE) when the two sides disagree, when the function blows up without bound, or when it oscillates forever. Keep these three failure modes in your back pocket.

“ $\lim$ exists” means the left-hand and right-hand limits are equal and finite. That single sentence settles most exam questions about existence.

Worked example 1

Estimate $\displaystyle\lim_{x\to 3}(2x-1)$ from the table.

1.From the left ( $x=2.9,\,2.99$ ) the outputs climb toward $5$ .
2.From the right ( $x=3.01,\,3.1$ ) the outputs fall toward $5$ .
3.Both sides agree, so the limit is $5$ . (Indeed $2(3)-1=5$ .)

Answer: $\displaystyle\lim_{x\to3}(2x-1)=5$

Worked example 2

The graph of $g$ has a hole at $x=1$ where the curve approaches height $4$ , but $g(1)=2$ (a filled dot). Find $\displaystyle\lim_{x\to1}g(x)$ and $g(1)$ .

1.The limit follows the approaching curve, which heads to $4$ .
2.The value $g(1)$ is read off the filled dot: $2$ .
3.They differ — which is perfectly allowed.

The limit and the function value are two separate questions. Mixing them up is the single most common Unit 1 error.

Answer: $\lim_{x\to1}g(x)=4$ , while $g(1)=2$ .

Worked example 3

Let $f(x)=\dfrac{x^2-4}{x-2}$ . Estimate $\displaystyle\lim_{x\to2}f(x)$ numerically.

1.At $x=2$ the formula gives $\frac{0}{0}$ : undefined, so there is a hole.
2.Try $x=1.99\!:\ \frac{1.99^2-4}{1.99-2}=\frac{-0.0399}{-0.01}=3.99$ .
3.Try $x=2.01\!:\ \approx 4.01$ . Both sides head to $4$ .

A $\frac00$ form does not mean the limit fails — it means you have work to do (Lesson 1.2).

Answer: $\displaystyle\lim_{x\to2}f(x)=4$

Worked example 4

For the piecewise function $f(x)=\begin{cases}x+1,& x<2\\ 5,& x\ge 2\end{cases}$ , does $\displaystyle\lim_{x\to2}f(x)$ exist?

1.Left side: as $x\to2^-$ , $f(x)=x+1\to 3$ .
2.Right side: as $x\to2^+$ , $f(x)=5$ .
3. $3\neq5$ , so the one-sided limits disagree.

Answer: The limit does not exist (jump).

Worked example 5

Explain why $\displaystyle\lim_{x\to0}\sin\!\left(\frac1x\right)$ does not exist.

1.As $x\to0$ , $\tfrac1x$ runs off to $\pm\infty$ .
2. $\sin$ of that swings between $-1$ and $1$ infinitely often, faster and faster.
3.The outputs never settle on one number.

Answer: DNE — infinite oscillation.

Worked example 6

From the graph, $\lim_{x\to a^-}f=1$ and $\lim_{x\to a^+}f=1$ , but there is an open hole at height $1$ and a filled dot at height $3$ . What is $\lim_{x\to a}f(x)$ ?

1.Both one-sided limits equal $1$ and agree.
2.The filled dot at $3$ is $f(a)$ ; it does not affect the limit.

Answer: $\lim_{x\to a}f(x)=1$ (even though $f(a)=3$ ).

Common trap: Reading the dot, not the approach

On a graph with an open circle at one height and a filled dot somewhere else, the limit follows the curve (the open circle), not the filled dot. The filled dot is $f(a)$ ; the curve tells you $\lim_{x\to a}f(x)$ . They are allowed to differ.

Hack: “Cover the dot”

Reading a limit off a graph? Put your fingertip over the open/filled dot at $x=a$ and trace the curve in from each side. Wherever your two fingers meet in midair is the limit. If they meet, the limit exists and equals that height; if they land at different heights, it’s DNE. The dot you covered never mattered.

What a Limit Is