Introduction
A line through the origin has the form \(f(x)=mx\), where \(m\) is a fixed real number. As the input \(x\) gets close to zero, the output \(mx\) also gets close to zero. The epsilon-delta definition lets us turn this intuitive observation into a precise proof: for any required output tolerance, we find an input distance that guarantees it.
The multiplier \(m\) may be positive, negative, or zero. The proof must work for every one of these possibilities. The key is to measure distance with absolute values, so the sign of the multiplier does not obscure the size of the output.
The Limit We Will Prove
Fix a real number \(m\), and define \(f(x)=mx\). The claim is that as \(x\) approaches zero, \(f(x)\) approaches zero. In limit notation, the claim is:
Here the input distance from zero is \(|x-0|=|x|\), and the output distance from the proposed limit zero is \(|mx-0|=|mx|\). The epsilon-delta definition therefore asks us to meet this condition:
The condition \(0<|x|\) means that \(x\) is near zero but is not equal to zero. A limit concerns values at nearby inputs, not just the value at the input being approached. In this example \(f(0)=0\), but the proof still establishes the required inequality for all sufficiently close nonzero inputs.
Why the Choice of Delta Depends on the Multiplier
The output error is \(|mx|\). The absolute-value product rule gives \(|mx|=|m||x|\). Thus the multiplier changes the size of the input distance by a factor of \(|m|\). If \(|m|\) is large, the input must be closer to zero to keep the output within the same tolerance. If \(|m|\) is small, a larger input distance can suffice.
When \(m\ne 0\), the natural choice is \(\delta=\varepsilon/|m|\). This makes the desired estimate follow directly from \(|x|<\delta\). When \(m=0\), division by \(|m|\) is not allowed, but the output error is exactly zero for every input, so any positive delta works.
Worked Examples
Worked Example: A Positive Multiplier
Prove from the definition that \(\lim_{x\to 0}3x=0\), using the specific tolerance \(\varepsilon=0.12\).
Because \(m=3\ne0\), choose \(\delta=\varepsilon/|m|\). Substituting the tolerance and multiplier gives:
Now suppose \(0<|x|<0.04\). Then:
Thus every nonzero input within distance \(0.04\) of zero produces an output within distance \(0.12\) of zero, as required. Checking the choice against the defining estimate gives \(3\delta=3(0.04)=0.12\); the strict inequality follows because the input satisfies \(|x|<\delta\), not merely \(|x|\le\delta\).
Worked Example: A Negative Multiplier
Prove from the definition that \(\lim_{x\to 0}(-5x)=0\), using \(\varepsilon=\frac{1}{10}\).
The multiplier has absolute value \(|-5|=5\). Choose:
Suppose \(0<|x|<\frac{1}{50}\). Regardless of whether \(x\) is positive or negative, absolute values give:
This verifies the defining condition. In particular, the negative sign in the formula does not make the output error negative: \(|-5x|\) is a nonnegative distance, equal to \(5|x|\).
Worked Example: The Zero Multiplier
Prove from the definition that \(\lim_{x\to 0}0x=0\), using \(\varepsilon=\frac{7}{10}\).
Here \(0x=0\) for every real input \(x\). Choose \(\delta=1\), which is positive. For any \(x\) satisfying \(0<|x|<1\), we have:
The conclusion is verified directly, and the particular positive delta was not important: any \(\delta>0\) would work. The formula \(\varepsilon/|m|\) cannot be used here because its denominator would be zero. The direct argument handles that case without division.
General Proof for Every Fixed Real Multiplier
We now prove the claim for an arbitrary fixed real number \(m\) and an arbitrary tolerance \(\varepsilon>0\). Since \(\varepsilon\) is arbitrary, the proof must provide a suitable positive \(\delta\) for every positive tolerance. Separate the cases \(m\ne0\) and \(m=0\) so that no division by zero occurs.
Let \(\varepsilon>0\) be given. We will choose a positive \(\delta\) that can depend on \(\varepsilon\) and on the fixed number \(m\), but not on \(x\).
Set \(\delta=\varepsilon/|m|\). Since \(\varepsilon>0\) and \(|m|>0\), this choice satisfies \(\delta>0\).
If \(0<|x|<\delta\), multiply \(|x|<\delta\) by the positive number \(|m|\). Then \(|mx|=|m||x|<|m|\delta=\varepsilon\).
Choose any \(\delta>0\), for example \(\delta=1\). For every \(x\) with \(0<|x|<\delta\), \(|mx|=|0x|=0<\varepsilon\).
In both cases, for every \(\varepsilon>0\), we have found a \(\delta>0\) such that \(0<|x|<\delta\) implies \(|mx|<\varepsilon\). This is exactly the epsilon-delta definition, so the limit is proved:
Why This Proof Works
The proof does not rely on checking a few values of \(x\). It controls every input satisfying the stated distance condition. Its central estimate is:
Once \(m\) is fixed, \(|m|\) is a fixed nonnegative number. When it is positive, restricting the input distance to less than \(\varepsilon/|m|\) ensures that multiplying that distance by \(|m|\) gives a number less than \(\varepsilon\). This is why the proof controls the output error, rather than merely observing that a few outputs appear small.
The sign of \(m\) is irrelevant to the size of the error because the definition measures distance using absolute value. A negative multiplier reverses the sign of the output, but it does not change the identity \(|mx|=|m||x|\). When \(m=0\), the output error vanishes altogether.
A Common Delta Mistake
A delta must be chosen to make the output error smaller than the requested tolerance. For example, consider the claim \(\lim_{x\to0}4x=0\) with \(\varepsilon=0.1\). A mistaken choice might be \(\delta=\varepsilon|m|=0.4\), multiplying by the slope instead of dividing by its absolute value. That choice does not guarantee the required estimate. Indeed, take \(x=0.3\). Then \(0<|x|=0.3<0.4=\delta\), but:
The correction is to choose \(\delta=\varepsilon/|m|\), which here gives \(\delta=0.1/4=0.025\). If \(0<|x|<0.025\), then:
As before, the final inequality is strict because \(|x|\) is strictly less than \(\delta\). Using \(\delta=\varepsilon/|m|\) creates the exact threshold; it does not mean that inputs at the boundary \(|x|=\delta\) must also satisfy the strict output inequality.
Summary
For a fixed real number \(m\), the output of \(f(x)=mx\) has distance \(|mx|=|m||x|\) from zero. If \(m\ne0\), choosing \(\delta=\varepsilon/|m|\) converts the input condition \(|x|<\delta\) into the required output condition \(|mx|<\varepsilon\). If \(m=0\), the output error is always zero and any positive delta works. Therefore the epsilon-delta definition proves \(\lim_{x\to0}mx=0\) for every fixed real multiplier.
Check Your Understanding
Use the definition and the estimates in this tutorial to answer the following questions.
- For \(m=6\) and \(\varepsilon=\frac{3}{20}\), what delta does the standard choice give? Verify the implication from \(0<|x|<\delta\) to \(|6x|<\varepsilon\).
- Why is \(\delta=\varepsilon/|m|\) not a valid formula when \(m=0\), and how is that case proved instead?
- For \(m=-2\), explain why the output error is \(2|x|\), not a negative quantity.
- In the general proof, why may the inequality \(|x|<\delta\) be multiplied by \(|m|\) when \(m\ne0\)?
- Does the proof require \(x\ne0\) to calculate \(|mx|=|m||x|\), or is that restriction part of the limit definition? Explain.
