Nearby Behavior and a Single Changed Value
A function assigns an output to each input in its domain. A limit asks a different question: as inputs get close to a particular number, what output value do the function values approach? The value assigned at the particular input is not what determines that nearby trend.
This distinction can be easy to miss on a graph. A single point may be moved far above or below the curve, yet the curve on either side can keep approaching exactly the same height. A table can show the same contrast: the row at the input may change, while the outputs at nearby inputs remain unchanged.
Function Value Versus Limit
The function value \(f(a)\) is the output assigned at the input \(a\). The limit \(\lim_{x\to a} f(x)=L\) describes a nearby pattern: when \(x\) is sufficiently close to \(a\), but \(x\ne a\), the value \(f(x)\) is close to \(L\).
Here is the precise condition we will use. For every positive tolerance \(\varepsilon\), there must be a positive distance \(\delta\) such that every input \(x\) satisfying \(0<|x-a|<\delta\) also satisfies \(|f(x)-L|<\varepsilon\). The condition deliberately excludes \(x=a\). It places no restriction on \(f(a)\).
The positive number \(\varepsilon\) specifies how close the output must be to \(L\); \(\delta\) specifies how close the input must be to \(a\). This statement does not say that every input closer to \(a\) must produce an output closer to \(L\) than every earlier input did. It says that, for each chosen output tolerance, all inputs within a suitable distance (except \(a\) itself) have outputs within that tolerance.
Changing a Point on a Graph
Suppose a curve approaches height \(4\) as \(x\) approaches \(2\). If we replace only the point at \(x=2\) with a point at height \(100\), the nearby parts of the curve do not move. The altered graph has a hole where the original point was, together with a filled point at a different height. The limit remains \(4\), even though the function value at \(2\) is now \(100\).
In graph language, the open circle indicates the height approached by the nearby curve, while the filled point indicates the function’s actual value at that input. The open circle is not itself the function value; it records the behavior of the curve around the input.
Worked Example: Changing the Value of \(x^2\) at \(x=2\)
Define \(g(x)=x^2\) for every real \(x\). Define \(f\) by \(f(x)=x^2\) when \(x\ne 2\), but let \(f(2)=100\). Compare their values near \(2\).
| Input \(x\) | \(g(x)=x^2\) | \(f(x)\) |
|---|---|---|
| 1.9 | 3.61 | 3.61 |
| 1.99 | 3.9601 | 3.9601 |
| 2 | 4 | 100 |
| 2.01 | 4.0401 | 4.0401 |
| 2.1 | 4.41 | 4.41 |
At every listed input other than \(2\), the two functions agree. At \(2\), they differ: \(g(2)=2^2=4\), while \(f(2)=100\). The entries on either side of \(2\) show outputs near \(4\), not near \(100\). Their graphs therefore share the same curve near \(x=2\), but have different filled points at \(x=2\).
For a precise check of the limit, let \(\varepsilon>0\) and choose \(\delta=\min(1,\varepsilon/5)\). If \(0<|x-2|<\delta\), then \(1<x<3\), so \(3<x+2<5\). Since \(x\ne2\), \(f(x)=x^2\), and hence
Thus \(\lim_{x\to2}f(x)=4\), even though \(f(2)=100\). The same calculation applies to \(g\), because \(g(x)=x^2\) near \(2\) as well. As a final check, the displayed table values come from direct substitution; for example, \(1.99^2=3.9601\) and \(2.01^2=4.0401\).
The Agreement-Nearby Principle
The example illustrates a general fact. Two functions do not have to agree everywhere to have the same limit at an input. It is enough for them to agree on some interval around that input with the input itself removed. Such an interval is called a deleted neighborhood of the input.
The theorem does not require \(f(a)\) and \(g(a)\) to be equal. In fact, either value can be different, or the function may not be defined at \(a\), without changing the reasoning. The hypothesis concerns only inputs different from \(a\).
Why the Agreement-Nearby Principle Works
Here is a proof using the precise limit condition above. Assume \(\lim_{x\to a}f(x)=L\), and suppose \(f(x)=g(x)\) whenever \(0<|x-a|<r\). We must show that \(g(x)\) also approaches \(L\).
Take any \(\varepsilon>0\). Since \(f\) has limit \(L\) at \(a\), there is a \(\delta_f>0\) such that \(0<|x-a|<\delta_f\) implies \(|f(x)-L|<\varepsilon\).
Set \(\delta=\min(\delta_f,r)\). If \(0<|x-a|<\delta\), then \(0<|x-a|<r\), so \(g(x)=f(x)\).
The same input also satisfies \(0<|x-a|<\delta_f\). Therefore \(|g(x)-L|=|f(x)-L|<\varepsilon\).
Because this works for every \(\varepsilon>0\), the precise limit condition holds for \(g\), and \(\lim_{x\to a}g(x)=L\). Notice that the proof never uses \(f(a)\) or \(g(a)\): the condition \(0<|x-a|\) excludes that input throughout.
Worked Examples with Tables and Exact Checks
Worked Example: A Simplified Expression with a Changed Point
Let \(g(x)=x+1\) for every real \(x\). Define \(h(x)=\frac{x^2-1}{x-1}\) when \(x\ne1\), and set \(h(1)=-7\). For \(x\ne1\), factor the numerator and cancel the nonzero factor \(x-1\):
The functions agree at every input except \(1\). The nearby values, along with the changed value, are:
| Input \(x\) | \(g(x)\) | \(h(x)\) |
|---|---|---|
| 0.9 | 1.9 | 1.9 |
| 0.99 | 1.99 | 1.99 |
| 1 | 2 | -7 |
| 1.01 | 2.01 | 2.01 |
| 1.1 | 2.1 | 2.1 |
The table suggests that both limits at \(1\) are \(2\). To check the claim for \(h\), take any \(\varepsilon>0\) and set \(\delta=\varepsilon\). Whenever \(0<|x-1|<\delta\), we have \(x\ne1\), so \(h(x)=x+1\). Thus
This verifies \(\lim_{x\to1}h(x)=2\). Substitution into \(g\) gives \(g(1)=1+1=2\), and the same inequality verifies \(\lim_{x\to1}g(x)=2\). The changed value \(h(1)=-7\) does not enter either limit check.
Worked Example: A Changed Point on a V-Shaped Graph
Define \(p(x)=|x-3|+2\) for every real \(x\). Let \(q(x)=p(x)\) for \(x\ne3\), but set \(q(3)=9\). For inputs near \(3\), the absolute value \(|x-3|\) measures the distance from \(x\) to \(3\).
| Input \(x\) | \(p(x)\) | \(q(x)\) |
|---|---|---|
| 2.9 | 2.1 | 2.1 |
| 2.99 | 2.01 | 2.01 |
| 3 | 2 | 9 |
| 3.01 | 2.01 | 2.01 |
| 3.1 | 2.1 | 2.1 |
For \(p\), the distance from the output to \(2\) is exactly the distance from the input to \(3\): \(|p(x)-2|=||x-3|+2-2|=|x-3|\). For \(q\), the same equality holds whenever \(x\ne3\). Given \(\varepsilon>0\), choose \(\delta=\varepsilon\). If \(0<|x-3|<\delta\), then \(x\ne3\) and
Therefore \(\lim_{x\to3}q(x)=2\). The identical calculation gives \(\lim_{x\to3}p(x)=2\). Direct substitution checks the exceptional values: \(p(3)=|0|+2=2\), while \(q(3)=9\). Again, the graph’s nearby V-shaped pieces approach height \(2\); only the filled point is moved.
A Common Mistake: Treating Closeness as a Pairwise Ranking
A limit does not mean that every input closer to \(a\) produces an output closer to \(L\) than every input farther away. Nearby outputs can fluctuate in their distances from \(L\). The requirement is instead that all outputs within a chosen input neighborhood meet a chosen output tolerance.
For instance, compare \(x=2.099\) and \(x=1.9\) for the function \(x^2\) near \(2\). Their input distances from \(2\) are \(0.099\) and \(0.1\), respectively, so \(2.099\) is closer to \(2\). But their output distances from \(4\) are
Thus the closer input has the larger output error in this comparison. The correct statement is the precise one: for every \(\varepsilon>0\), there is a \(\delta>0\) such that \(0<|x-2|<\delta\) implies \(|x^2-4|<\varepsilon\). This condition concerns all inputs inside a suitable neighborhood; it does not impose a ranking on each pair of inputs.
Why This Distinction Matters
Limits describe local behavior, while a function value records what the function does at one specific input. Keeping those roles separate makes it possible to interpret graphs with holes, compare formulas that agree except at one point, and distinguish an approached value from an assigned value.
A limit can exist even if the function is not defined at the input, or if the function’s value there differs from the limit. Continuity at \(a\), by contrast, requires that \(f(a)\) be defined, that the limit as \(x\) approaches \(a\) exist, and that the two values agree. So changing a point may leave the limit unchanged while changing whether the function is continuous there.
Check Your Understanding
Use the distinction between the function value and nearby behavior to answer these questions.
- Suppose \(f(x)=x^2\) for \(x\ne3\), but \(f(3)=20\). What is \(\lim_{x\to3}f(x)\), and what is \(f(3)\)?
- In the precise limit condition, why is the requirement \(0<|x-a|\) included?
- Two functions agree whenever \(0<|x-4|<0.1\), but have different values at \(4\). What can you conclude about their limits at \(4\), if one of those limits exists?
- For \(r(x)=|x-1|+5\), what is the distance \(|r(x)-5|\) in terms of \(x\)?
- Does a limit equal to \(L\) guarantee that the function value at the input is \(L\)? Explain briefly.
