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Calculator Strategy and Error Control · Tutorial 984 of 999

Reading Numerical Tables for Limits and Change

Use carefully chosen inputs on both sides of a target to read numerical evidence for limits and local function change, while recognizing what a table cannot establish.

Intermediate 9 min read

What You'll Learn

  • Choose input values that approach a target from the left and right.
  • Read a table to estimate a finite limit when the function may be undefined at the target.
  • Compare one-sided table patterns to detect a possible two-sided limit failure.
  • Track changes in input and output separately when observing local function behavior.
  • Explain why even a detailed finite table cannot prove that a limit exists.

Why Read a Table Near a Target?

A numerical table can make behavior near a particular input easier to see. By entering values that get closer to a target from both sides, you can observe whether the corresponding function values appear to approach a common number, move in different directions, or behave irregularly.

This is useful when a graph is hard to read or when a formula is difficult to evaluate mentally. But a table is evidence, not a proof: it contains only finitely many sampled inputs, and its displayed values may be rounded.

Key idea: To estimate a two-sided limit at \(x=a\), inspect function values for inputs just less than \(a\) and just greater than \(a\). A common pattern on both sides suggests a limit value, but finite table entries alone cannot establish that the limit exists.

Build the Table on Both Sides

Suppose you want to estimate \(\lim_{x\to a} f(x)\). A useful table uses inputs approaching \(a\) from below and above. For example, values such as \(a-0.1\), \(a-0.01\), \(a+0.01\), and \(a+0.1\) sample both sides. The inputs should be close enough to reveal local behavior, but not so close that calculator precision, cancellation, or rounding obscures the pattern.

Keep the input \(x=a\) separate from the nearby values. A limit concerns what happens as \(x\) approaches \(a\), not necessarily the value \(f(a)\). The function may be undefined at \(a\), or its value there may differ from the limiting value.

When reading a table, look for a pattern rather than simply selecting the closest row. Ask whether the function values from the left and right appear to settle near the same number. If they appear to approach different values, the two-sided limit does not exist. If they appear to approach the same value, the table supports that estimate but does not prove it.

Worked Example: Estimate a Limit at a Missing Input

Worked Example: Estimate a Limit at a Missing Input

Use a table to estimate \(\lim_{x\to 1} \frac{x^2-1}{x-1}\), and explain what the function does at \(x=1\).

The original expression is undefined at \(x=1\), because its denominator is zero there. For \(x\ne1\), factor the numerator and simplify:

$$ \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1. $$

Use inputs on each side of 1. The table values follow from \(x+1\), which agrees with the original expression whenever \(x\ne1\).

Input \(x\)Function valueSide of 1
0.91.9Left
0.991.99Left
0.9991.999Left
1.0012.001Right
1.012.01Right
1.12.1Right

For example, at \(x=0.999\), the simplified expression gives \(0.999+1=1.999\), matching the table value. The table values are correct. From both sides, the function values approach 2, so the numerical estimate is 2. At \(x=1\), however, the original function is undefined.

Check the estimate using the simplified expression. As \(x\) gets close to 1, \(x+1\) gets close to 2. The simplification is valid for nearby inputs other than 1, exactly the inputs used to evaluate the limit.

Why the Table Pattern Is Reasonable

The preceding example has a short algebraic justification. For \(x\ne1\), the original function equals \(x+1\). The difference between that value and the proposed limit 2 is:

$$ |(x+1)-2|=|x-1|. $$

Thus, whenever \(x\ne1\) is within \(\varepsilon\) of 1, the function value is within \(\varepsilon\) of 2. More precisely, for any \(\varepsilon>0\), choose \(\delta=\varepsilon\). If \(0<|x-1|<\delta\), then the denominator is nonzero and

$$ \left|\frac{x^2-1}{x-1}-2\right| = |x-1| < \varepsilon. $$

This proves that the limit is 2. The table suggests the result; the algebraic argument verifies it for every sufficiently close input, not just the sampled ones.

Limit versus function value: A limit describes nearby function values. It does not require \(f(a)\) to exist, and it does not require \(f(a)\) to equal the limit.

Compare the Two Sides

A two-sided limit can exist only if the function values approach the same number from the left and from the right. These one-sided patterns should therefore be inspected separately before making an overall estimate. A combined list of values can conceal an important difference between the two sides.

In a calculator table, label or otherwise track which inputs are below the target and which are above it. If the values on one side seem to settle near one number and those on the other side near another, do not average them. Different one-sided limits mean there is no two-sided limit.

Worked Example: Detect Different One-Sided Behavior

Worked Example: Detect Different One-Sided Behavior

For \(f(x)=\frac{x-3}{|x-3|}\), use a table to decide whether \(\lim_{x\to3} f(x)\) exists.

At \(x=3\), the denominator is zero, so the function is undefined. If \(x<3\), then \(x-3<0\) and \(|x-3|=-(x-3)\), giving \(f(x)=-1\). If \(x>3\), then \(|x-3|=x-3\), giving \(f(x)=1\).

Input \(x\)CalculationFunction value
2.9\((-0.1)/0.1\)\(-1\)
2.99\((-0.01)/0.01\)\(-1\)
2.999\((-0.001)/0.001\)\(-1\)
3.001\(0.001/0.001\)\(1\)
3.01\(0.01/0.01\)\(1\)
3.1\(0.1/0.1\)\(1\)

The left-side values are all \(-1\), while the right-side values are all 1. In fact, the formulas show that the left-hand limit is \(-1\) and the right-hand limit is 1. Since these are different, the two-sided limit does not exist. The table’s pattern is consistent with the exact calculation.

Check one entry from each side: \(f(2.999)=(-0.001)/0.001=-1\), and \(f(3.001)=0.001/0.001=1\). These calculations confirm that the displayed values follow the function’s definition.

What a Finite Table Cannot Establish

Even a table with many rows samples only finitely many inputs. A function might behave differently at inputs the table did not test. This is why matching values near a target can support a limit estimate but cannot, by itself, prove that the limit exists.

The following example makes the limitation precise. Define a function near 2 by setting \(g(x)=1\) when \(x=2+\frac1n\) for some positive integer \(n\), and \(g(x)=0\) at all other inputs. Also set \(g(2)=0\).

Worked Example: A Table That Misses Nearby Exceptions

Worked Example: A Table That Misses Nearby Exceptions

Suppose a calculator table samples \(x=1.91,\ 1.991,\ 1.9991,\ 2.09,\ 2.009,\ 2.0009\). None of these is 2 plus the reciprocal of a positive integer, so \(g(x)=0\) at every sampled input. The table appears to suggest that the limit is 0.

But consider inputs \(x_n=2+\frac1n\). They get arbitrarily close to 2 as \(n\) increases, and \(g(x_n)=1\) for every positive integer \(n\). For instance, \(x_{100}=2.01\) and \(g(2.01)=1\). Also consider \(y_n=2+\frac{1}{n+1/2}\). These inputs approach 2, but they are not of the form \(2+\frac1m\) for a positive integer \(m\): equality would require \(m=n+\frac12\), which is not an integer. Therefore \(g(y_n)=0\) for every positive integer \(n\).

Along inputs approaching 2, the function can consequently take the value 1 or the value 0 arbitrarily close to the target. It cannot approach one common limit. The finite table missed the exceptional inputs, so its apparent pattern did not establish a limit.

Check the exceptional input: \(x_{100}=2+1/100=2.01\), and the definition gives \(g(2.01)=1\). The listed sample \(2.009\) differs from 2 by \(0.009\), which is not \(1/n\) for a positive integer \(n\), so its value is 0 as stated.

Use Tables to Observe Function Change

A table can also help you describe how much a function’s output changes as its input changes near a target. Keep the input change and output change distinct. For a chosen target \(a\), compare \(f(x)\) with \(f(a)\); the output change is \(f(x)-f(a)\), while the input change is \(x-a\). This is an observation about function values, not a conclusion that every nearby input behaves exactly like the sampled ones.

1
Choose the target.
Identify the input \(a\) and calculate \(f(a)\), if it is defined.
2
Sample both sides.
Choose inputs a little below and above \(a\), then evaluate the function.
3
Compare changes.
Compute \(x-a\) and \(f(x)-f(a)\) to see how the output varies near the target.
4
State the evidence carefully.
Describe the pattern shown by the table without treating the samples as proof of all nearby behavior.

Worked Example: Track Output Changes Near a Target

Worked Example: Track Output Changes Near a Target

For \(h(x)=x^2\), use a table near \(x=3\) to observe the change in function value from \(h(3)\).

First calculate \(h(3)=3^2=9\). For each nearby input, calculate \(h(x)\) and then subtract 9.

\(x\)Input change \(x-3\)\(h(x)=x^2\)Output change \(h(x)-9\)
2.9\(-0.1\)\(2.9^2=8.41\)\(8.41-9=-0.59\)
2.99\(-0.01\)\(2.99^2=8.9401\)\(8.9401-9=-0.0599\)
3.01\(0.01\)\(3.01^2=9.0601\)\(9.0601-9=0.0601\)
3.1\(0.1\)\(3.1^2=9.61\)\(9.61-9=0.61\)

The table shows that the output changes are smaller for the sampled inputs closer to 3. For example, moving from 3 to 2.99 changes the input by \(-0.01\) and the output by \(-0.0599\); moving from 3 to 3.01 changes them by \(0.01\) and \(0.0601\), respectively. The differences are not identical on the two sides, so report the observed values rather than forcing a symmetric pattern.

Check the output changes directly: \(8.9401-9=-0.0599\) and \(9.0601-9=0.0601\). Both entries match the table. This comparison describes the sampled function change; it does not turn a finite list of calculations into a general theorem.

Common Pitfall: Treating the Nearest Row as the Answer

A frequent error is to use only the input closest to the target and declare its function value to be the limit. For the first example, seeing \(x=0.999\) gives the value \(1.999\). It would be incorrect to conclude that the limit is \(1.999\): that is the function value at one nearby input, not the value approached as inputs get arbitrarily close to 1.

The correction is to inspect a sequence of increasingly close inputs on both sides and identify the value the function values appear to approach. Here, \(1.9,\ 1.99,\ 1.999\) produce \(1.9,\ 1.99,\ 1.999\), while inputs just above 1 produce values just above 2. The pattern suggests 2, and the factorization proves it. Keep the conclusion proportional to the evidence: tables estimate; mathematical reasoning establishes.

Check Your Understanding

Use these questions to check how you would interpret and report numerical evidence.

  1. When estimating a two-sided limit at \(x=a\), why should a table include inputs on both sides of \(a\)?
  2. If a function is undefined at \(a\), can its limit as \(x\to a\) still exist? Explain using the first worked example.
  3. For \(f(x)=\frac{x-3}{|x-3|}\), what do the table values show about the two-sided limit at 3?
  4. Why does the table in the example involving \(g\) fail to prove that the limit is 0?
  5. For \(h(x)=x^2\), state the input change and output change when \(x\) moves from 3 to 3.01.

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