From a Linear Formula to a Limit Proof
A line changes at a constant rate: when its input changes by a given amount, its output changes by the slope times that amount. This simple relationship makes linear functions especially suitable for learning how an epsilon-delta proof is built. The proof still requires care: a limit claim is not established just by guessing the answer or choosing a delta without checking the implication.
In this tutorial, we will prove limits for affine functions, which have the form \(f(x)=mx+b\), where \(m\) and \(b\) are fixed real numbers. The coefficient \(m\) is the slope. The central calculation is that the difference between the function value and its proposed limit is exactly the absolute value of the slope times the distance from the input to the specified point.
The Definition and Its Domain Condition
The epsilon-delta definition describes how close \(f(x)\) must be to a proposed limit \(L\) whenever \(x\) is sufficiently close to a point \(a\), but not equal to \(a\). The domain condition matters: the definition applies when \(a\) is an accumulation point of the domain. This means every open interval around \(a\) contains a domain point different from \(a\).
Here, \(\varepsilon\) is the permitted error in the output and \(\delta\) is a restriction on the input distance. The strict inequalities are part of the definition. In particular, the proof must establish an output error strictly less than \(\varepsilon\) from an input distance strictly less than \(\delta\).
For an affine function defined on all of \(\mathbb R\), every real number \(a\) is an accumulation point of the domain: for every positive radius, there are real numbers other than \(a\) within that distance. Thus the domain condition is satisfied at each input considered below. The value of a function at \(a\) itself does not enter the punctured condition \(0<|x-a|<\delta\).
The Error Calculation That Guides the Choice
For \(f(x)=mx+b\), the natural candidate for the limit as \(x\) approaches \(a\) is the function value \(ma+b\). To see why, subtract this candidate from \(f(x)\): the constant term cancels, and the remaining difference factors by \(x-a\). Taking absolute values turns the slope into its magnitude. This identity explains how to select \(\delta\).
If \(m\ne 0\), making \(|x-a|\) less than \(\varepsilon/|m|\) makes the output error less than \(\varepsilon\). If \(m=0\), the function is constant and the output error is zero for every input; any positive \(\delta\) works. In the examples with nonzero slope, we will use the direct choice \(\delta=\varepsilon/|m|\).
Worked Examples
Worked Example: A Positive Slope at a Nonzero Input
Prove that \(\lim_{x\to 3}(4x-7)=5\). Let \(f(x)=4x-7\), so the proposed limit is \(4(3)-7=12-7=5\). Let \(\varepsilon>0\) be arbitrary. Choose \(\delta=\varepsilon/4\), which is positive because \(\varepsilon>0\).
Now take any real \(x\) satisfying \(0<|x-3|<\delta\). Then
Thus every \(x\) in the required punctured neighborhood has \(|f(x)-5|<\varepsilon\). Since this works for every positive \(\varepsilon\), the definition gives \(\lim_{x\to 3}(4x-7)=5\).
Check: The candidate value is \(4(3)-7=5\), as used in the proof. The error calculation reduces exactly to \(4|x-3|\), and the chosen \(\delta\) makes that quantity strictly less than \(\varepsilon\).
Worked Example: A Negative Slope and a Concrete Tolerance
Prove that \(\lim_{x\to 2}(-3x+4)=-2\). First evaluate the affine expression at the specified input: \(-3(2)+4=-6+4=-2\). For a concrete tolerance, take \(\varepsilon=0.06\) and choose \(\delta=\varepsilon/3=0.06/3=0.02\).
Suppose \(0<|x-2|<0.02\). Then
This verifies the required implication for the stated tolerance. To establish the full limit, repeat the same argument for arbitrary \(\varepsilon>0\): choose \(\delta=\varepsilon/3\), and then \(0<|x-2|<\delta\) gives
Therefore \(\lim_{x\to 2}(-3x+4)=-2\). The negative sign of the slope disappears only after taking the absolute value; the error is nonnegative and equals \(3|x-2|\).
Check: Substitution into the formula gives \(-3(2)+4=-2\), and for the concrete choice \(\delta=0.02\), the error is strictly less than \(0.06\) whenever the input distance is strictly less than \(0.02\).
Worked Example: A Fractional Slope
Prove that \(\lim_{x\to -1}\left(\frac{1}{2}x+3\right)=\frac{5}{2}\). The proposed value is \(\frac{1}{2}(-1)+3=-\frac{1}{2}+3=\frac{5}{2}\). Let \(\varepsilon>0\). Since the slope has magnitude \(\frac{1}{2}\), choose \(\delta=\varepsilon/(\frac{1}{2})=2\varepsilon\), which is positive.
If \(0<|x-(-1)|<\delta\), then
This proves the limit by the definition. Notice that the input restriction is larger than \(\varepsilon\) here, because the slope has magnitude less than one: input changes produce smaller output changes.
Check: The candidate value is \(\frac{5}{2}\), and the exact output error is \(\frac{1}{2}|x+1|\). With \(\delta=2\varepsilon\), this error is strictly less than \(\varepsilon\), as required.
The General Linear Limit Theorem
The theorem follows directly from the error identity, but a complete proof must verify the quantified definition. In particular, it must begin with an arbitrary positive \(\varepsilon\), give a positive \(\delta\), and show that every domain input satisfying the punctured-neighborhood condition has the required output error.
Let \(\varepsilon>0\) be arbitrary. We will produce a \(\delta>0\) that works for this tolerance.
If \(m\ne 0\), set \(\delta=\varepsilon/|m|\). This is positive because both \(\varepsilon\) and \(|m|\) are positive. If \(m=0\), set \(\delta=1\), which is positive.
Assume \(x\in\mathbb R\) and \(0<|x-a|<\delta\). If \(m\ne 0\), use the exact error identity to obtain \(|f(x)-(ma+b)|=|m|\,|x-a|<|m|\delta=\varepsilon\). If \(m=0\), then \(f(x)=b\) and \(ma+b=b\), so \(|f(x)-(ma+b)|=|b-b|=0<\varepsilon\).
For every \(\varepsilon>0\), the selected positive \(\delta\) makes the required implication true for every \(x\in\mathbb R\). Since \(a\) is an accumulation point of \(\mathbb R\), the definition applies and establishes the stated limit.
Why the Choice Works—and What It Does Not Say
The choice of \(\delta\) is not a guess detached from the proof. It comes from solving the error estimate in the direction needed. The exact relationship is \(|f(x)-(ma+b)|=|m|\,|x-a|\). To make the left side less than \(\varepsilon\), it is enough to make \(|x-a|\) less than \(\varepsilon/|m|\) when \(m\ne0\). The strict inequality in the hypothesis then produces a strict inequality in the conclusion.
A valid \(\delta\) need not be the largest possible choice, and it need not be unique. Any positive \(\delta\) that makes the implication hold is acceptable. The formula \(\varepsilon/|m|\) is convenient because it follows immediately from the exact error identity. It also makes clear that a steeper line requires a smaller input neighborhood for the same output tolerance.
The proof establishes a limit, not a claim about a measured or rounded value. It also does not require \(f(a)\) to be considered separately from nearby function values: for this affine function, \(f(a)=ma+b\), but the defining implication concerns \(x\ne a\). The accumulation-point condition prevents the definition from becoming vacuous when there are no domain points near \(a\) other than \(a\) itself.
Common Pitfall: Dropping the Absolute Value
A frequent error is to treat a negative output difference as automatically small. For example, consider \(f(x)=-2x+5\) as \(x\) approaches \(1\). The candidate limit is \(-2(1)+5=3\). A mistaken attempt might claim that \(\delta=\varepsilon\) works because \(f(x)-3=-2(x-1)<\varepsilon\). This inequality alone is not the required conclusion: the definition asks for \(|f(x)-3|<\varepsilon\).
The failure is concrete when \(\varepsilon=0.1\). The incorrect choice \(\delta=0.1\) permits \(x=1.075\), because \(0<|1.075-1|=0.075<0.1\). But
The correction is to keep the absolute value and account for the slope magnitude. Choose \(\delta=\varepsilon/2=0.05\). Then \(0<|x-1|<0.05\) implies
The issue was not that a negative slope prevents a proof. It was that the proof must control the magnitude of the output difference, regardless of its sign.
A Reliable Proof Checklist
When writing a linear epsilon-delta proof at a specified input, keep the order of reasoning explicit:
- Evaluate the affine expression at \(a\) to identify the proposed limit \(ma+b\).
- Begin with an arbitrary \(\varepsilon>0\).
- Choose a positive \(\delta\), using the slope magnitude when \(m\ne0\).
- Assume \(x\) is in the domain and \(0<|x-a|<\delta\).
- Compute \(|f(x)-(ma+b)|\) and show each step leading to a quantity strictly less than \(\varepsilon\).
- State that the definition gives the limit; ensure the specified input is an accumulation point of the domain.
This sequence separates the choice of \(\delta\) from its verification. Both are essential: a plausible choice without a checked implication is not a complete proof, and an estimate without a positive \(\delta\) does not fulfill the definition.
Check Your Understanding
For each question, focus on the proposed limit, the delta choice, and the implication that must be verified.
- For \(f(x)=5x-1\) as \(x\) approaches \(2\), what is the proposed limit, and what delta choice works for arbitrary \(\varepsilon>0\)?
- For \(f(x)=-4x+7\) as \(x\) approaches \(-1\), write the error \(|f(x)-L|\) in terms of \(|x-(-1)|\), and give a valid delta.
- Why does a proof for a negative slope require an absolute value around the output difference?
- If \(f(x)=6\) for all real \(x\), explain why any positive \(\delta\) works for the proposed limit \(6\).
- What condition on the domain is required at the input \(a\) for the epsilon-delta definition of a limit to apply?
