Verifying a Proposed Delta

What Does It Mean to Verify a Proposed Delta?

A proposed value of $\delta$ is useful only if it does what the epsilon-delta condition requires: every input sufficiently close to $a$ must produce an output within the requested tolerance $\varepsilon$ of the proposed limit $L$. Verifying a delta means checking this implication using an estimate for the output error.

Definition: For a function $f$, a point $a$, a proposed limit $L$, and a tolerance $\varepsilon>0$, a number $\delta>0$ guarantees the requested bound if, whenever $0<|x-a|<\delta$, we have $|f(x)-L|<\varepsilon$.

The strict inequalities matter. The input must be closer than $\delta$, not exactly $\delta$ away, and the output error must be strictly less than $\varepsilon$. In the linear and quadratic examples here, the proposed limit is $L=f(a)$. We will check whether a stated positive $\delta$ guarantees the requested bound by substituting it into an estimate for $|f(x)-f(a)|$.

Verifying a Delta for a Linear Function

Let $f(x)=mx+b$, where $m$ and $b$ are constants. For any input $x$ and point $a$, subtracting the function values gives

$$ f(x)-f(a)=(mx+b)-(ma+b)=m(x-a). $$

Taking absolute values shows that the output error is the slope magnitude multiplied by the input distance. If $m\ne 0$ and $|x-a|<\delta$, then

|f(x)-f(a)|=|m||x-a|<|m|\delta.

Thus, a proposed $\delta$ passes this test whenever $|m|\delta\le\varepsilon$. Notice that equality is allowed in this last comparison: the output error is strictly less than $|m|\delta$, so it is still strictly less than $\varepsilon$ when $|m|\delta=\varepsilon$.

Worked Example: Checking a Delta for a Linear Function

Consider $f(x)=3x-2$ at $a=1$, with proposed limit $L=f(1)$, tolerance $\varepsilon=0.12$, and proposed $\delta=0.04$. First calculate the limit value:

$$ f(1)=3(1)-2=1. $$

For any $x$, the output error is

$$ |f(x)-f(1)| =|(3x-2)-1| =|3x-3| =3|x-1|. $$

If $0<|x-1|<0.04$, then

$$ |f(x)-1|=3|x-1|<3(0.04)=0.12. $$

The computed upper bound is exactly $\varepsilon$, but the error is strictly less than that bound. Therefore this $\delta$ guarantees $|f(x)-1|<0.12$. Checking the proposed value in the criterion gives $|m|\delta=3(0.04)=0.12=\varepsilon$, as required.

If $m=0$, the function is constant: $f(x)=b$. Then $f(x)-f(a)=0$ for every $x$, and $0<\varepsilon$ for every positive tolerance. Every positive $\delta$ therefore works. This case should be stated separately rather than using a strict estimate whose right-hand side is zero.

Turning a Quadratic Input Restriction into an Error Bound

Now let $f(x)=Ax^2+Bx+D$, where $A$, $B$, and $D$ are constants. The constant term cancels when we subtract $f(a)$. Set $h=x-a$, so $x=a+h$. Factoring the difference gives

\begin{aligned} f(x)-f(a) &=A(x^2-a^2)+B(x-a)\\ &=(x-a)\bigl(A(x+a)+B\bigr)\\ &=h\bigl(2Aa+B+Ah\bigr). \end{aligned}

Write $C=2Aa+B$. If $|x-a|=|h|<\delta$, the triangle inequality gives the useful estimate

|f(x)-f(a)| =|h||C+Ah| \le |h|\bigl(|C|+|A||h|\bigr).

When $(A,B)\ne(0,0)$, the quantity $|C|+|A|\delta$ is positive. Indeed, if $A\ne0$, its second term is positive; if $A=0$, then $B\ne0$ and $C=B\ne0$. For $h\ne0$, the restrictions $|h|<\delta$ and $|h|<\delta$ in the two factors therefore imply the strict estimate

|f(x)-f(a)| <\delta\bigl(|2Aa+B|+|A|\delta\bigr).

If $h=0$, the output error is zero, so the requested strict epsilon bound holds directly. This also covers the excluded input $x=a$, if one chooses to include it in the implication. If $A=B=0$, the function is constant, and its output error is zero; handle that case directly instead of applying the displayed strict estimate with a zero right-hand side.

Test for a nonconstant quadratic: For $f(x)=Ax^2+Bx+D$, with $(A,B)\ne(0,0)$, a proposed $\delta>0$ guarantees $|f(x)-f(a)|<\varepsilon$ whenever $$ \delta\bigl(|2Aa+B|+|A|\delta\bigr)\le\varepsilon. $$ The test follows by placing the proposed $\delta$ into the error estimate. The comparison may be equality because the estimate for the output error is strict.

Worked Quadratic Verifications

Worked Example: A Quadratic Delta That Passes

Let $f(x)=2x^2-3x+1$, $a=1$, $\varepsilon=0.3$, and proposed $\delta=0.1$. The proposed limit is

$$ f(1)=2(1)^2-3(1)+1=2-3+1=0. $$

Here $A=2$ and $B=-3$, so the coefficient at $a=1$ is

$$ 2Aa+B=2(2)(1)+(-3)=4-3=1. $$

Substitute the proposed delta into the estimate:

\delta\bigl(|2Aa+B|+|A|\delta\bigr) =0.1\bigl(|1|+|2|(0.1)\bigr) =0.1(1+0.2) =0.12.

Since $0.12\le0.3$, the estimate proves the requested bound. In detail, if $0<|x-1|<0.1$, then

$$ |f(x)-f(1)| <0.12 <0.3. $$

Thus the proposed $\delta=0.1$ guarantees the output error is less than $0.3$. Substitution into the test confirms the comparison $0.12\le0.3$.

Worked Example: Equality in the Quadratic Test

Take $f(x)=x^2$, $a=0$, $\varepsilon=\frac14$, and proposed $\delta=\frac12$. The limit value is $f(0)=0$. The coefficients are $A=1$, $B=0$, so $2Aa+B=0$. Substitution gives

\delta\bigl(|2Aa+B|+|A|\delta\bigr) =\frac12\left(0+1\cdot\frac12\right) =\frac14 =\varepsilon.

Although the calculated bound equals $\varepsilon$, the output-error estimate is strict. Whenever $0<|x|<\frac12$,

|f(x)-f(0)|=|x^2|=|x|^2 <\left(\frac12\right)^2 =\frac14.

Therefore the proposed delta works. Substituting it back into the error expression verifies the conclusion: the error is strictly below $\frac14$, not merely at most $\frac14$.

Worked Example: A Constant Quadratic

Let $f(x)=5$, written as $Ax^2+Bx+D$ with $A=0$, $B=0$, and $D=5$. At $a=2$, the proposed limit is $f(2)=5$. For any positive $\varepsilon$ and any positive proposed $\delta$,

|f(x)-f(2)|=|5-5|=0<\varepsilon.

So, for example, $\varepsilon=0.01$ and $\delta=3$ satisfy the required implication for every $x$. This check is direct. The nonconstant-quadratic estimate does not apply here: its right-hand side would be $3(0+0)=0$, and one must not claim $0<0$.

Why the Estimate Works

The method is an instance of a simple proof pattern: begin with the output error, rewrite it so the input distance appears as a factor, and then use the assumed input restriction to bound the factors. For a quadratic, the factorization shows exactly where the dependence on $x-a$ occurs. The triangle inequality controls the other factor without needing to know the particular input $x$.

Start with the output error.
Subtract $f(a)$ from $f(x)$ and take the absolute value.

Factor out the input difference.
For the quadratic, $f(x)-f(a)=(x-a)(2Aa+B+A(x-a))$.

Apply the input restriction.
Use $|x-a|<\delta$ to replace the varying distance in the estimate by the proposed delta.

Compare with the tolerance.
If the resulting bound is at most $\varepsilon$, the strict output-error estimate gives the required error less than $\varepsilon$.

This reasoning establishes a sufficient check: when the proposed delta makes the displayed upper bound at most $\varepsilon$, the implication is proved. If a particular estimate does not establish that comparison, it has not verified the delta by this route. Do not silently replace an unsuccessful estimate with a claim that the delta works; any such claim needs its own justification.

A Common Strict-Inequality Mistake

A common mistake is to reject a proposed delta whenever the computed upper bound equals $\varepsilon$. For $f(x)=x^2$ at $a=0$, with $\delta=\frac12$ and $\varepsilon=\frac14$, the mistaken calculation is: “The bound is $\frac14$, and $\frac14$ is not less than $\frac14$, so the delta fails.” That compares the bound with $\varepsilon$ but overlooks the strict inequality that precedes it.

0<|x|<\frac12 \quad\Longrightarrow\quad |x^2|=|x|^2<\left(\frac12\right)^2=\frac14=\varepsilon.

The correction is that the estimate gives output error strictly less than the calculated bound. A bound equal to $\varepsilon$ is therefore enough. A separate mistake is to use the nonconstant-quadratic strict estimate when $A=B=0$; in that constant case, verify $0<\varepsilon$ directly.

Check Your Understanding

For each question, identify the proposed limit as the function value at the given point, then use the relevant error estimate.

For $f(x)=4x+1$ at $a=2$, does $\delta=0.05$ guarantee an error below $\varepsilon=0.2$? Show the comparison.
For $f(x)=-2x+3$, what output-error estimate follows from $|x-a|<\delta$?
For $f(x)=x^2+2x$ at $a=1$, calculate $|2Aa+B|+|A|\delta$ when $\delta=0.1$.
Why can a proposed delta pass the test when the calculated bound equals $\varepsilon$?
For $f(x)=7$, why should the constant case be checked directly rather than by the strict nonconstant-quadratic estimate?

What You'll Learn

What Does It Mean to Verify a Proposed Delta?

Verifying a Delta for a Linear Function

Worked Example: Checking a Delta for a Linear Function

Turning a Quadratic Input Restriction into an Error Bound

Worked Quadratic Verifications

Worked Example: A Quadratic Delta That Passes

Worked Example: Equality in the Quadratic Test

Worked Example: A Constant Quadratic

Why the Estimate Works

A Common Strict-Inequality Mistake

Check Your Understanding

Test yourself: Calculus I final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

Verifying a Proposed Delta

What You'll Learn

What Does It Mean to Verify a Proposed Delta?

Verifying a Delta for a Linear Function

Worked Example: Checking a Delta for a Linear Function

Turning a Quadratic Input Restriction into an Error Bound

Worked Quadratic Verifications

Worked Example: A Quadratic Delta That Passes

Worked Example: Equality in the Quadratic Test

Worked Example: A Constant Quadratic

Why the Estimate Works

A Common Strict-Inequality Mistake

Check Your Understanding

Test yourself: Calculus I final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

Previous

Next