From an Output Tolerance to an Input Tolerance
A limit statement asks us to control the output of a function by restricting how close the input is to a specified point. The desired output accuracy is represented by \(\varepsilon>0\). We seek a positive input tolerance \(\delta\) such that whenever \(x\) is within \(\delta\) of \(a\), but \(x\ne a\), the function value is within \(\varepsilon\) of the proposed limit \(L\).
The key task in this tutorial is to derive a suitable \(\delta\) from the inequality we need to satisfy. The value of \(\delta\) may depend on \(\varepsilon\), on the function, and on the point \(a\). It must be positive, and the resulting implication must hold for every input meeting its condition. There is usually more than one valid choice: making \(\delta\) smaller preserves the implication.
The Basic Strategy
Start with the output error, \(|f(x)-L|\), and rewrite or bound it in terms of \(|x-a|\). Then solve the resulting inequality for the input distance. If the bound only holds near \(a\), include that local restriction when choosing \(\delta\). Finally, check that the chosen formula is positive and that it implies the required strict output inequality.
Begin with \(|f(x)-L|<\varepsilon\), where \(\varepsilon>0\) is given.
Factor or simplify the expression, and use a local bound if needed.
Solve the resulting inequality and combine it with any restriction needed for the bound.
Use \(0<|x-a|<\delta\) to show the output error is strictly less than \(\varepsilon\).
Linear and Affine Functions
For an affine function \(f(x)=mx+b\), the value at \(a\) is \(ma+b\). Subtracting that value cancels the constant term. If \(m\ne0\), the error is exactly a constant multiple of the input distance. This makes the choice of \(\delta\) direct.
To make this less than \(\varepsilon\), it is enough to require \(|x-a|<\varepsilon/|m|\). Because \(\varepsilon>0\) and \(|m|>0\), this choice is positive. If \(m=0\), the function is constant and its error is zero for every input; any positive \(\delta\) works.
Worked Example: Choosing Delta for an Affine Function
Consider \(f(x)=-2x+5\) as \(x\) approaches \(1\). The proposed limit is \(f(1)=-2+5=3\). Let the desired output tolerance be \(\varepsilon=1/3\).
First compute the error:
We want \(2|x-1|<1/3\), so it is enough to require \(|x-1|<1/6\). Thus choose \(\delta=1/6\), which is positive.
To check the choice, suppose \(0<|x-1|<1/6\). Then
Therefore this \(\delta\) gives the required output accuracy.
The Square Function at Zero
For the limit of \(x^2\) at zero, the output error is \(|x^2-0|=|x|^2\). Since both \(|x|\) and \(\varepsilon\) are nonnegative, requiring \(|x|<\sqrt{\varepsilon}\) makes its square less than \(\varepsilon\). The square root is positive when \(\varepsilon>0\), so this gives a positive choice of \(\delta\).
Worked Example: Choosing Delta for the Square at Zero
Suppose we want \(|x^2|<1/16\) whenever \(x\) is sufficiently close to zero. Here \(\varepsilon=1/16\), so choose
This is positive. If \(0<|x-0|<1/4\), then
Substituting the chosen \(\delta\) into the error bound gives the required strict inequality, so the choice works.
Quadratic Limits at an Arbitrary Input
For a limit of \(x^2\) at an arbitrary real number \(a\), the difference factors as \((x-a)(x+a)\). The second factor is not a fixed constant, so we first restrict \(x\) to a fixed neighborhood of \(a\). If \(|x-a|<1\), the triangle inequality gives
Consequently, throughout this neighborhood we have the non-strict bound
The non-strict inequality is valid even at \(x=a\), where both sides are zero. To ensure that the right-hand side is less than \(\varepsilon\), it is enough to require \(|x-a|<\varepsilon/(2|a|+1)\). We also need \(|x-a|<1\) for the local bound. Both restrictions can be met by taking their minimum.
Why the Quadratic Choice Works
The formula follows from the factorization and the local bound. This argument also checks the positivity and the strict output inequality required by the definition.
Since \(\varepsilon>0\) and \(2|a|+1\ge1\), both \(1\) and \(\varepsilon/(2|a|+1)\) are positive. Their minimum \(\delta\) is therefore positive.
If \(0<|x-a|<\delta\), then \(|x-a|<1\) and \(|x-a|<\varepsilon/(2|a|+1)\), because \(\delta\) is no larger than either bound.
The factorization and local estimate give \(|x^2-a^2|\le(2|a|+1)|x-a|<(2|a|+1)\varepsilon/(2|a|+1)=\varepsilon\).
The strict inequality at the end comes from \(|x-a|<\varepsilon/(2|a|+1)\). The factor estimate itself need not be strict after multiplication; keeping track of these two different inequalities avoids a common logical error.
Worked Example: Choosing Delta for a Quadratic Limit at a Negative Input
Consider \(\lim_{x\to-1}x^2=1\), and let \(\varepsilon=1/2\). Here \(a=-1\), so \(2|a|+1=3\). The general choice gives
This value is positive. Now suppose \(0<|x-(-1)|=|x+1|<1/6\). In particular, \(|x+1|<1\), so
Therefore the error satisfies
Thus every input within the chosen distance, except the limit point itself as specified by the definition, has output error less than \(\varepsilon\).
A Common Mistake: Losing the Local Restriction
The quadratic estimate depends on keeping \(x\) within distance \(1\) of \(a\). Without that restriction, the factor \(|x+a|\) cannot be bounded by the same fixed number. For example, at \(a=2\), the claim \(|x+2|\le4\) whenever \(|x-2|<1\) is false: \(x=5/2\) satisfies \(|x-2|=1/2<1\), but \(|x+2|=9/2>4\).
The correct estimate uses the triangle inequality:
It follows that \(|x^2-4|\le5|x-2|\) under this restriction. Writing a strict inequality \(|x^2-4|<5|x-2|\) for every \(x\) with \(|x-2|<1\) would be incorrect at \(x=2\), because both sides are then zero. The non-strict bound is valid there and is sufficient: the separate condition \(|x-2|<\delta\) supplies the strict inequality needed at the final step.
Another tempting shortcut is to choose a value from the error inequality but forget a restriction needed to justify the bound used to obtain it. For a quadratic, a complete choice must satisfy both the local-distance requirement and the output-tolerance requirement. Taking their minimum handles both at once.
Choosing Does Not Mean There Is Only One Answer
A formula for \(\delta\) is a useful sufficient choice, not necessarily the only choice. If a positive \(\delta\) works, then any smaller positive value also works: the condition \(0<|x-a|<\delta_{\text{small}}\) restricts \(x\) to a subset of the inputs covered by the original choice. In particular, a derivation may give a convenient bound rather than the largest possible input tolerance.
The central habit is to work from the error toward the input distance while retaining every condition used along the way. For affine functions the error is an exact multiple of \(|x-a|\). For the square function at zero it is \(|x|^2\). For a general quadratic point, factorization creates a second factor that must first be bounded locally. In each case, the proposed \(\delta\) must be positive and the final estimate must be strictly less than \(\varepsilon\).
Check Your Understanding
For each question, identify the inequality that connects the output error to the input distance and state any restriction needed for it.
- For \(f(x)=4x-3\) at \(a=2\), what choice of \(\delta\) follows from \(\varepsilon=1/5\)?
- For \(x^2\) at zero, what choice of \(\delta\) follows from \(\varepsilon=1/49\)?
- For \(x^2\) at \(a=3\), write the local bound for \(|x^2-a^2|\) when \(|x-a|<1\).
- Why is \(\min(1,\varepsilon/(2|a|+1))\) positive whenever \(\varepsilon>0\)?
- Why is it safer to use a non-strict inequality for the quadratic error bound that includes \(x=a\)?
