Calculus I · Limits and Continuity · lesson

Limits at Holes and Undefined Points

A function can have a limit where it is undefined

Consider

f(x)=x24x2.f(x)=\frac{x^2-4}{x-2}.

At x=2x=2, the formula is undefined. For x2x\ne2, however,

f(x)=(x2)(x+2)x2=x+2.f(x)=\frac{(x-2)(x+2)}{x-2}=x+2.

So the graph is the line y=x+2y=x+2 with one point removed.

Reference table
xxf(x)f(x)
1.93.9
1.993.99
1.9993.999
2.0014.001
2.014.01
2.14.1

Both sides approach 44, so

limx2x24x2=4.\boxed{\lim_{x\to2}\frac{x^2-4}{x-2}=4}.
Graph of y equals x plus 2 with a removable hole at (2, 4).
Read this graph as text

A limit at a removable hole. The line y = x + 2 is drawn on two explicit domains, one to the left of 2 and one to the right. An open circle at (2, 4) marks the missing value. Dashed horizontal and vertical guides identify x = 2 and y = 4, so the common approached height remains clear without relying on color.

The curve is split into left and right branches and the missing value is an open circle with dashed coordinate guides.

Why it matters: Show that a two-sided limit depends on nearby outputs even when the function is undefined at the target input.

Read the graph

The function is undefined at x=2x=2, but nearby outputs approach 44.

Method

To estimate a limit from a table, use inputs on both sides of the target. The table should move progressively closer to the target. A table gives evidence, not proof; algebra or a trustworthy graph is usually needed to confirm the pattern.

Worked example

Table, graph, and algebra agree

Estimate and then evaluate

limx1x31x1.\lim_{x\to1}\frac{x^3-1}{x-1}.
Show worked solution

Numerical evidence.

Reference table
xxx31x1\dfrac{x^3-1}{x-1}
0.90.92.712.71
0.990.992.97012.9701
0.9990.9992.9970012.997001
1.0011.0013.0030013.003001
1.011.013.03013.0301
1.11.13.313.31

The values appear to approach 33.

Algebraic confirmation. Use the difference-of-cubes identity:

x31=(x1)(x2+x+1).x^3-1=(x-1)(x^2+x+1).

For x1x\ne1,

x31x1=x2+x+1.\frac{x^3-1}{x-1}=x^2+x+1.

Therefore,

limx1x31x1=limx1(x2+x+1)=12+1+1=3.\begin{aligned} \lim_{x\to1}\frac{x^3-1}{x-1} &=\lim_{x\to1}(x^2+x+1)\\ &=1^2+1+1\\ &=\boxed{3}. \end{aligned}

After the explanation

Use the section idea

Reading lens

What are nearby outputs doing as the input approaches the target from both sides?

Mental model

Imagine tightening a window around the target input and watching where all nearby outputs are forced to gather.

Decision

Read the left-hand and right-hand behavior separately first; combine them only after both sides approach the same output.

Common trap

The function value at the target can be missing or deliberately moved, so never substitute a plotted dot for evidence from both sides.

Check yourself

You understand the section when you can explain a limit from a graph, table, and sentence without confusing it with the function value.

Source & rights

Original instruction with traceable references.

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The verified handoff declares original composition and requires owner provenance review. BetterGrades-original material remains separate from public-domain references; no source textbook PDF is published here.

Vocab
Limit
Math glossaryLimit
limxaf(x)=L\lim_{x\to a}f(x)=L

The value a function approaches as its input approaches a target.

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Function
Math glossaryFunction
y=f(x)y=f(x)

A rule or relation assigning exactly one output to each allowed input.

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Continuity
Math glossaryContinuity
limxaf(x)=f(a)\lim_{x\to a}f(x)=f(a)

At a point, the function value exists and equals the limit there.

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Domain
Math glossaryDomain
domf\operatorname{dom}f

The set of permitted input values.

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Math glossary