Three Kinds of Evidence for a Limit
A graph, a table, and an epsilon-delta argument can all help us investigate a limit, but they do not do the same job. A graph can make a pattern visible. A table can show what happens at selected inputs from both sides. An epsilon-delta argument can establish a precise claim for every input close enough to the point.
It is useful to compare these forms of evidence rather than ask which one is always best. In practice, a graph or table can help identify a plausible limit or reveal a problem. Exact reasoning is what confirms the claim under the definition of a limit.
The definition describes a two-sided limit: inputs may approach \(a\) from below or above. It says that any requested positive tolerance \(\varepsilon\) for the output can be met by choosing a suitable positive distance \(\delta\) for the input. A graph and a table can illustrate that idea, but checking only a finite number of inputs does not check the condition for every input in a whole interval.
What Each Kind of Evidence Establishes
| Evidence | What it is useful for | What it cannot establish by itself |
|---|---|---|
| Graph | Suggesting the value approached from either side and spotting visible jumps or holes. | Exact behavior at every sufficiently close input; a plotted graph has limited resolution. |
| Two-sided table | Checking selected values below and above the input and comparing their outputs. | Behavior at all nearby inputs; finitely many rows are not an epsilon-delta proof. |
| Epsilon-delta argument | Proving a limit claim, or using a contradiction to rule one out, for all relevant nearby inputs. | It may be less useful than a picture for first noticing a pattern, and it must be logically correct. |
For a two-sided limit, the evidence from the left and right must be considered together. If the function appears to approach different values on the two sides, a two-sided limit cannot exist. If a table displays similar values on both sides, that supports a candidate, but it does not by itself rule out different behavior between the sampled inputs.
Worked Example: A Hole and a Different Function Value
Worked Example: A Hole and a Different Function Value
Define \(f(x)=2x+1\) when \(x\ne 2\), and define \(f(2)=10\). Compare the graph, a table, and an exact argument to investigate \(\lim_{x\to 2}f(x)\).
Graphical evidence. For every \(x\ne 2\), the graph follows the line \(y=2x+1\). That line approaches \(5\) as \(x\) approaches \(2\). The graph therefore has an open circle at \((2,5)\). Since \(f(2)=10\), it also has a filled point at \((2,10)\). The graph suggests a limit of \(5\), even though the function value is \(10\).
Tabular evidence. Use inputs on both sides of \(2\). At each listed input, \(x\ne 2\), so \(f(x)=2x+1\).
The outputs below \(2\) are below \(5\), and those above \(2\) are above \(5\). In this small sample, they get closer to \(5\) as the inputs get closer to \(2\). That is evidence for the candidate \(L=5\), not yet a proof.
Exact evidence. Let \(\varepsilon>0\) be arbitrary. Choose \(\delta=\varepsilon/2\), which is positive. Suppose \(0<|x-2|<\delta\). In particular, \(x\ne 2\), so the formula \(f(x)=2x+1\) applies. Then $$ |f(x)-5|=|2x+1-5|=|2x-4|=2|x-2|<2\delta=\varepsilon. $$ Thus every positive \(\varepsilon\) has a corresponding positive \(\delta\) that meets the definition, so \(\lim_{x\to 2}f(x)=5\).
Check. The same calculation shows the output error is exactly twice the input distance whenever \(x\ne 2\): \(|f(x)-5|=2|x-2|\). The choice \(\delta=\varepsilon/2\) therefore gives the required strict inequality. The separate value \(f(2)=10\) does not enter the limit condition, because that condition requires \(0<|x-2|\).
Why the Exact Argument Is Stronger Than a Sample
The proof in the first example works because it starts with an arbitrary tolerance and controls the output error for every input satisfying the chosen distance condition. Its conclusion is not limited to the four table entries. This is the central difference between checking examples and proving a limit.
The structure of the reasoning is worth making explicit. We first fix an arbitrary \(\varepsilon>0\). We then choose a positive \(\delta\) based on that \(\varepsilon\). Finally, we assume \(0<|x-a|<\delta\) and show that \(|f(x)-L|<\varepsilon\). The choice must work for all such \(x\), not just for a few convenient values.
Look for a plausible output value and compare behavior on both sides of the input.
Name the input \(a\) and the proposed limit \(L\).
Apply the epsilon-delta definition, or show that it cannot be satisfied.
Worked Example: A Quadratic Limit from Both Sides
Worked Example: A Quadratic Limit from Both Sides
Investigate whether \(\lim_{x\to 1}x^2=1\) using a graph, a table, and an exact argument.
Graphical evidence. The graph \(y=x^2\) is a parabola. Near \(x=1\), its values approach \(1\) from inputs less than \(1\) and from inputs greater than \(1\). The graph suggests the two-sided limit is \(1\).
Tabular evidence. Calculate values on both sides:
For example, the output errors at \(0.99\) and \(1.01\) are \(1-0.9801=0.0199\) and \(1.0201-1=0.0201\). The sampled outputs on each side are close to \(1\) when the inputs are close to \(1\). Again, these particular calculations suggest, but do not prove, the limit.
Exact argument. Let \(\varepsilon>0\) be arbitrary, and choose \(\delta=\min(1,\varepsilon/3)\). This is positive. If \(0<|x-1|<\delta\), then \(|x-1|<1\), so \(0<x<2\). Therefore \(1<x+1<3\), and hence \(|x+1|<3\). Now factor the difference of squares:
The final inequality follows from \(\delta\leq\varepsilon/3\). Thus the epsilon-delta definition is satisfied, and \(\lim_{x\to 1}x^2=1\).
Check. The bound \(|x+1|<3\) was justified by first ensuring \(|x-1|<1\). Without that restriction, the factor \(|x+1|\) would not have the stated bound for every real \(x\). The chosen \(\delta\) provides both the needed input restriction and the required error tolerance.
When the Two Sides Disagree
A graph or table may reveal that the outputs approach different values from the left and the right. That is not merely an uncertainty about which candidate to choose: it rules out a two-sided limit. The exact definition can show why, by demonstrating that no single \(L\) can meet its requirements.
Worked Example: Different One-Sided Behavior
Define \(g(x)=x/|x|\) for \(x\ne 0\), and set \(g(0)=0\). Determine whether \(g\) has a two-sided limit as \(x\) tends to \(0\).
Graphical evidence. For \(x<0\), \(g(x)=-1\); for \(x>0\), \(g(x)=1\). The graph has a horizontal branch at \(-1\) to the left of \(0\), and a horizontal branch at \(1\) to the right. The value \(g(0)=0\) is a separate point. The two sides suggest incompatible candidate limits.
Tabular evidence. Values from either side are constant:
The table shows the disagreement clearly for these inputs. To rule out a limit, we need an argument that applies regardless of how close the inputs are chosen.
Exact argument. Suppose, for contradiction, that the two-sided limit exists and equals some real number \(L\). Take \(\varepsilon=1/2\). The definition would give a \(\delta>0\) such that every \(x\) with \(0<|x|<\delta\) satisfies \(|g(x)-L|<1/2\). Choose \(x=\delta/2\) and \(z=-\delta/2\). Both are nonzero, and $$ |x|=\delta/2<\delta,\qquad |z|=\delta/2<\delta. $$ So the definition would imply \(|1-L|<1/2\) and \(|-1-L|<1/2\). By the triangle inequality, $$ 2=|1-(-1)|\leq |1-L|+|L-(-1)|<\frac12+\frac12=1, $$ which is impossible. Therefore no such \(L\) exists, and \(g\) has no two-sided limit at \(0\).
Check. The contradiction does not depend on the value assigned at \(x=0\), since the definition considers only \(0<|x|<\delta\). It rules out every proposed \(L\), not just \(L=0\), \(L=1\), or \(L=-1\).
A Common Mistake: Treating a Table as a Proof
Suppose a table gives outputs close to a proposed limit for several inputs on both sides. It is a mistake to conclude that the limit has been proved simply because the listed values are close. The definition requires the output condition to hold for all inputs in a punctured interval around the point. A finite table checks only its listed inputs.
For instance, the table in the first example supports the claim that the limit is \(5\), but it does not establish that every input sufficiently close to \(2\) has an output sufficiently close to \(5\). The correction is to use the exact relation \(|f(x)-5|=2|x-2|\), then choose \(\delta\) in terms of an arbitrary \(\varepsilon\). The table helps discover the claim; the argument proves it.
There is a related visual pitfall: a graph may look smooth because a plotting window cannot display a narrow jump or rapid change. A graph is valuable for forming expectations and checking whether a proposed conclusion is plausible, but the scale and resolution of a drawing are not part of the limit definition.
Using the Evidence Together
A careful limit investigation often moves from an informal view to an exact conclusion. First, inspect the graph for behavior on both sides and note any hole or separately defined point. Next, calculate a small two-sided table to test the candidate and catch arithmetic or interpretation errors. Then use the epsilon-delta definition to establish the claim or to show that no single limit can work.
Agreement among a graph, a table, and an exact proof is useful because each checks the question in a different way. But the roles are not interchangeable: visual and numerical evidence guide the investigation, while the exact argument determines whether the definition is satisfied.
Check Your Understanding
For each question, distinguish evidence that suggests a limit from reasoning that establishes it.
- Why should a table include inputs on both sides of the point when investigating a two-sided limit?
- A graph has an open circle at \((a,L)\) and a filled point at \((a,M)\), where \(L\ne M\). What does this suggest about the limit and the function value?
- Why can a finite table not, by itself, prove that a limit exists?
- In the proof for \(x^2\) at \(1\), why was it useful to require \(\delta\leq 1\) as well as \(\delta\leq\varepsilon/3\)?
- How does the contradiction in the example for \(g(x)=x/|x|\) show that no real number can be its two-sided limit at \(0\)?
