Introduction
For an affine function, the output changes in direct proportion to a change in the input. That simple relationship makes its limits especially accessible to an epsilon-delta proof: once we specify how close the output must be to the proposed limit, the slope tells us how close the input must be to its target.
Here we will prove the limit for \(f(x)=mx+b\) as \(x\) approaches an arbitrary real number \(a\). The proof will cover positive, negative, and zero slopes, and will show exactly how to choose a suitable input tolerance.
The Claim and Its Key Estimate
Let \(m\) and \(b\) be fixed real numbers, and define \(f(x)=mx+b\). At the input \(a\), the function has value \(ma+b\). The proposed limit is therefore the function’s value at that input. To prove this claim, compare the output at \(x\) with the proposed limit:
This equality is the central estimate. The constant term cancels, and the magnitude of the output error is the input error multiplied by \(|m|\). Using the absolute value matters because the output error is a nonnegative distance whether the slope is positive or negative.
The epsilon-delta definition says that \(\lim_{x\to a} f(x)=L\) if, for every \(\varepsilon>0\), there is a \(\delta>0\) such that whenever \(0<|x-a|<\delta\), we have \(|f(x)-L|<\varepsilon\). The values \(\varepsilon\) and \(\delta\) measure the required output and input tolerances, respectively.
Why the Slope Determines Delta
Suppose first that \(m\ne 0\). To make the output error less than \(\varepsilon\), the key estimate shows that it is enough to make \(|m|\,|x-a|<\varepsilon\). Dividing by the positive number \(|m|\), we see that it is enough to require \(|x-a|<\varepsilon/|m|\). This gives a positive choice of delta for every positive epsilon.
If \(m=0\), then \(f(x)=b\) for every input. Its output error from \(b\) is always zero, so any positive delta works. The proof below treats both cases explicitly.
The proof works because it provides a direct guarantee: inputs within the chosen distance of \(a\) produce outputs within the requested distance of \(ma+b\). The restriction \(0<|x-a|\) in the definition excludes the input \(x=a\), but the estimate itself also holds at \(x=a\).
Worked Example: A Positive Slope
Worked Example: A Positive Slope
Prove directly that \(\lim_{x\to 4}(3x-2)=10\). Here \(m=3\), \(b=-2\), and \(a=4\), so the proposed limit is \(ma+b=3(4)-2=10\).
Let \(\varepsilon>0\) be given. Since \(|m|=3\), choose \(\delta=\varepsilon/3\), which is positive. If \(0<|x-4|<\delta\), then
This verifies the definition for every positive \(\varepsilon\). For example, if the required output tolerance is \(\varepsilon=0.06\), then \(\delta=0.06/3=0.02\). An input satisfying \(0<|x-4|<0.02\) has output error \(3|x-4|<3(0.02)=0.06\). Thus the proposed limit is verified.
Worked Example: A Negative Slope
Worked Example: A Negative Slope
Prove directly that \(\lim_{x\to 1}(-2x+5)=3\). Substituting the target input into the affine expression gives \((-2)(1)+5=3\).
Let \(\varepsilon>0\) be given. The slope is \(m=-2\), so \(|m|=2\). Choose \(\delta=\varepsilon/2\). For any \(x\) satisfying \(0<|x-1|<\delta\), calculate the output error:
Therefore the epsilon-delta condition holds. For a concrete check, take \(\varepsilon=0.1\). The chosen input tolerance is \(\delta=0.1/2=0.05\), and whenever \(0<|x-1|<0.05\), the output error is \(2|x-1|<2(0.05)=0.1\). The negative slope reverses the direction of change, but it does not make the distance negative.
Worked Example: A Zero Slope
Worked Example: A Zero Slope
Prove that \(\lim_{x\to -3}(0x+7)=7\). The expression is the constant function \(f(x)=7\), and the proposed limit is \(0(-3)+7=7\).
Let \(\varepsilon>0\) be given. Choose \(\delta=1\), a positive number. If \(0<|x-(-3)|<1\), then \(f(x)=7\), and hence
This verifies the definition. The choice \(\delta=1\) is not special: any positive delta would work, because the output error is zero for every input. In this case the formula \(\varepsilon/|m|\) is unavailable because \(|m|=0\); that is why the zero-slope case is handled separately.
A Reusable Proof Structure
For a nonzero slope, the proof can be organized around the same few decisions each time. First identify the proposed limit by evaluating the affine expression at \(a\). Then express the output error as \(|m|\,|x-a|\), and choose \(\delta=\varepsilon/|m|\). Finally, use the hypothesis \(|x-a|<\delta\) to conclude that the output error is less than \(\varepsilon\).
For \(f(x)=mx+b\), the proposed value at \(a\) is \(ma+b\).
Simplify \(|f(x)-(ma+b)|\) to \(|m|\,|x-a|\).
If \(m\ne 0\), set \(\delta=\varepsilon/|m|\). If \(m=0\), take any positive \(\delta\), such as \(1\).
Use \(0<|x-a|<\delta\) to show \(|f(x)-(ma+b)|<\varepsilon\).
Common Pitfall: Forgetting the Absolute Value of the Slope
A frequent error is to divide by \(m\) rather than by \(|m|\). This can produce a negative value for delta, even though the definition requires \(\delta>0\). For example, with slope \(m=-2\) and tolerance \(\varepsilon=0.1\), the incorrect choice \(\delta=\varepsilon/m=0.1/(-2)=-0.05\) is not an allowable input tolerance.
The correction is to use the error in distance, which is nonnegative: \(|-2|=2\). Thus \(\delta=\varepsilon/|m|=0.1/2=0.05>0\). If \(0<|x-a|<0.05\), then the output error is \(2|x-a|<2(0.05)=0.1\). The absolute value handles negative slopes correctly and ensures that the chosen delta is positive.
Another useful check is to substitute the chosen delta into the estimate before concluding. For nonzero \(m\), the calculation must end with \(|m|\delta=\varepsilon\), with a strict inequality before that final bound. This makes clear both why the choice works and where the strict output tolerance comes from.
Why This Result Matters
This proof illustrates a basic strategy for limit arguments: simplify the output error until it can be controlled by the input distance. In this case, the relationship is exact, not merely an upper bound. The slope measures how strongly an input change is amplified in the output, so a larger \(|m|\) requires a proportionally smaller delta for the same epsilon.
The result also confirms that an affine function has no jump or gap in its limiting value at any real input: its limit is the value obtained by substituting the target input. The justification, however, comes from the epsilon-delta argument rather than from substitution alone. The argument explicitly shows that every requested output tolerance can be met by a suitable input tolerance.
Check Your Understanding
Use the error relation between input and output to answer these questions.
- For \(f(x)=5x+1\) as \(x\to 2\), what is the proposed limit?
- For a nonzero slope \(m\), what choice of \(\delta\) guarantees \(|m|\,|x-a|<\varepsilon\) whenever \(|x-a|<\delta\)?
- Why must the absolute value \(|m|\), rather than \(m\), appear in the delta formula?
- What is a suitable delta when the affine function has slope zero, and why does it work?
- If \(m=-4\) and \(\varepsilon=0.2\), what delta does the proof choose?
