Why Input Choice Matters
A table can only show what happens at the inputs you choose. If every input is far from the target, the table may give little information about nearby behavior. If the inputs are poorly arranged, it may also be difficult to tell which values are closer to the target and from which side they approach it.
A useful limit table therefore uses inputs that are near the target, on both sides, and progressively closer. The target itself is usually left out: a limit concerns nearby inputs, and the function might not even be defined at the target.
This tutorial focuses on constructing the input column. The function values can then be calculated for those inputs, as in the earlier material on estimating limits from tables.
Build Inputs from Distances
Suppose the target is \(c\). An input is to the left of \(c\) when it is less than \(c\), and to the right when it is greater than \(c\). A convenient way to choose inputs is to decide on a list of positive distances, such as \(0.1\), \(0.01\), and \(0.001\).
For each distance \(h\), form one input on each side:
For instance, if \(c=2\) and \(h=0.01\), then the corresponding inputs are \(2-0.01=1.99\) and \(2+0.01=2.01\). Both are exactly \(0.01\) from the target, but they lie on opposite sides.
A typical set of distances decreases by a factor of ten. This gives a straightforward pattern of inputs that are successively closer to the target:
There is no requirement that every table use these exact distances. The target’s location and the function’s domain matter. The important features are that the distances are positive, become smaller, and produce valid inputs.
Why These Inputs Get Closer
The rule works because the distance from a real number \(x\) to \(c\) is \(|x-c|\). If \(x=c-h\), where \(h>0\), then \(x\) is left of \(c\), and its distance from \(c\) is \(h\). If \(x=c+h\), then \(x\) is right of \(c\), and its distance is also \(h\).
Therefore, if \(0<h_2<h_1\), the inputs made with \(h_2\) are closer to \(c\) than the corresponding inputs made with \(h_1\). For example, \(0.01<0.1\), so \(c-0.01\) is closer to \(c\) than \(c-0.1\), and \(c+0.01\) is closer than \(c+0.1\). This verifies the input-selection rule directly: the chosen distance is exactly the distance from the input to the target.
Worked Example: Inputs on Both Sides of 2
Worked Example: Inputs on Both Sides of 2
Construct inputs on both sides of \(c=2\), using distances \(0.1\), \(0.01\), and \(0.001\). Then make a table for \(f(x)=x^2+1\). Do not include \(x=2\).
For the left side, subtract each distance from 2:
For the right side, add each distance to 2:
Now substitute each chosen input into \(f(x)=x^2+1\). For example, at \(x=1.9\),
At \(x=1.99\) and \(x=1.999\), the calculations are
On the right side, the substitutions give
The resulting table puts the inputs closest to 2 nearest to the target row, while keeping the two sides separate:
| Side | Input \(x\) | Distance from 2 | \(f(x)=x^2+1\) |
|---|---|---|---|
| Left | 1.9 | 0.1 | 4.61 |
| Left | 1.99 | 0.01 | 4.9601 |
| Left | 1.999 | 0.001 | 4.996001 |
| Right | 2.001 | 0.001 | 5.004001 |
| Right | 2.01 | 0.01 | 5.0401 |
| Right | 2.1 | 0.1 | 5.41 |
Check the closest left and right rows by substituting them again: \((1.999)^2+1=3.996001+1=4.996001\), and \((2.001)^2+1=4.004001+1=5.004001\). These agree with the table. The input values are correctly placed because \(1.999<2<2.001\), and each is \(0.001\) from 2.
Keep the Target Out of the Input List
It is easy to confuse the target with the nearby inputs. In a table intended to examine values near \(c\), the row \(x=c\) is not another nearby sample: it is the target itself. Including it is unnecessary for choosing inputs and can be impossible if the function is undefined there.
For example, consider \(f(x)=\dfrac{x^2-9}{x-3}\) near \(c=3\). The denominator is zero when \(x=3\), so \(f(3)\) is undefined. Inputs such as \(2.9\) and \(3.1\), however, are valid. They can be selected using distances from 3 without evaluating the function at 3.
Worked Example: A Function Undefined at the Target
Worked Example: A Function Undefined at the Target
Choose table inputs for \(f(x)=\dfrac{x^2-9}{x-3}\) near \(c=3\), using distances \(0.1\), \(0.01\), and \(0.001\). Calculate the function values without including \(x=3\).
Subtracting the distances gives the left-side inputs \(2.9\), \(2.99\), and \(2.999\). Adding them gives the right-side inputs \(3.1\), \(3.01\), and \(3.001\). All six differ from 3, so the denominator \(x-3\) is nonzero at each input.
To calculate the values efficiently, factor the numerator:
For any \(x\ne 3\), cancellation gives
This cancellation is valid for the six chosen inputs because none equals 3. Substituting each input into \(x+3\) gives the table entries:
| Side | Input \(x\) | Distance from 3 | \(f(x)\) |
|---|---|---|---|
| Left | 2.9 | 0.1 | \(2.9+3=5.9\) |
| Left | 2.99 | 0.01 | \(2.99+3=5.99\) |
| Left | 2.999 | 0.001 | \(2.999+3=5.999\) |
| Right | 3.001 | 0.001 | \(3.001+3=6.001\) |
| Right | 3.01 | 0.01 | \(3.01+3=6.01\) |
| Right | 3.1 | 0.1 | \(3.1+3=6.1\) |
Check against the original quotient using \(x=2.9\):
For the closest right-side input, the check is
Both checks agree with the table. The target itself was not used: the quotient has no value at \(x=3\).
Choose Distances That Fit the Target
The same method works when the target is negative, a decimal, or both. First identify the target carefully, then add and subtract the distances. Do not assume that the target is zero or that the input values should simply be \(0.1\), \(0.01\), and \(0.001\). Those are distances; the actual inputs depend on \(c\).
Worked Example: A Negative Decimal Target
Construct a table for \(f(x)=(x+1.5)^2+2\) near \(c=-1.5\), using distances \(0.1\) and \(0.01\). Omit the target input.
For the left side, subtract each distance from \(-1.5\):
For the right side, add each distance:
Evaluate the function at each input. At \(x=-1.6\), for example,
The other values are
| Side | Input \(x\) | Distance from \(-1.5\) | \(f(x)\) |
|---|---|---|---|
| Left | -1.6 | 0.1 | 2.01 |
| Left | -1.51 | 0.01 | 2.0001 |
| Right | -1.49 | 0.01 | 2.0001 |
| Right | -1.4 | 0.1 | 2.01 |
Check the inputs by measuring their distances: \(|-1.51-(-1.5)|=|-0.01|=0.01\), and \(|-1.49-(-1.5)|=|0.01|=0.01\). Substitution checks the corresponding values: \((\pm0.01)^2+2=0.0001+2=2.0001\). The target \(-1.5\) is not among the inputs.
A Practical Selection Process
Before calculating function values, use a short process to make sure the input column is useful and valid.
Write down the input \(c\) that the table is meant to approach.
Start with manageable increments, for example \(0.1\), \(0.01\), and \(0.001\), or choose other distances suitable for the target.
For each distance \(h\), calculate \(c-h\) and \(c+h\).
Confirm that left inputs are less than \(c\), right inputs are greater than \(c\), and every selected input is in the function’s domain.
Do not include \(c\) as one of the nearby inputs. Then evaluate the function at the remaining inputs.
Common Mistake: Using the Distance as the Input
A frequent error is to use the chosen increments themselves as inputs, even when the target is not zero. Suppose the target is \(c=5\) and the distance is \(h=0.1\). The inputs are not \(0.1\) and \(-0.1\). Those numbers are not \(0.1\) from 5.
The incorrect calculation is:
For example, the distance from \(-0.1\) to 5 is \(|-0.1-5|=5.1\), not \(0.1\). Use the target plus or minus the distance instead:
Now the distance checks work: \(|4.9-5|=0.1\) and \(|5.1-5|=0.1\). Writing down \(c-h\) and \(c+h\) before evaluating the function helps prevent this mistake.
Making the Table Useful
Use a consistent set of distances on the two sides when you want the table to be easy to compare. Keep enough decimal places to distinguish the chosen inputs. For example, \(3.001\) and \(3.01\) are different inputs, even though their digits are similar. If a calculator displays values rounded, retain sufficient precision for the table rather than rounding the inputs prematurely.
Always check the domain before calculating. A function may be defined on only part of the real line, so an input that is on the correct side of \(c\) may still be invalid. In that case, choose a smaller or otherwise suitable distance that produces allowed inputs. The goal is not to fill a table mechanically; it is to select valid inputs whose distances from the target are clear.
Check Your Understanding
For each question, focus on the input choices before calculating any function values.
- For target \(c=4\), write the left and right inputs at distances \(0.1\) and \(0.01\). Order each side from farther to closer.
- For target \(c=-2.5\), what inputs are \(0.001\) away on the left and right?
- Why should a table of nearby inputs usually omit the target itself?
- A learner uses \(0.01\) and \(-0.01\) as inputs for a target of \(c=3\). What is the mistake, and what are the correct inputs at distance \(0.01\)?
- For \(f(x)=\dfrac{x^2-16}{x-4}\) near \(c=4\), are inputs \(3.99\) and \(4.01\) valid? Explain using the function’s domain.
