Why Compare the Two Approaches?
A secant slope measures the average rate of change between two points on a function. In the earlier material on shrinking intervals, the moving point approached a fixed point and we tracked how the slope changed. There is an important refinement: the moving point can approach from the left or from the right, and those two approaches need not behave alike.
To make the distinction precise, fix an input \(a\). One point on the graph is \((a,f(a))\); the other is \((x,f(x))\), where \(x\ne a\). The secant slope depends on whether \(x<a\) or \(x>a\). Comparing these cases helps us see whether the nearby behavior is consistent on both sides.
The expression \(S_a(x)\) is not defined at \(x=a\), because its denominator would be zero. We study what its values do for nearby inputs other than \(a\). A left-hand limit \(L\) means that for every \(\varepsilon>0\), there is a \(\delta>0\) such that \(0<a-x<\delta\) implies \(|S_a(x)-L|<\varepsilon\). A right-hand limit is defined similarly, with \(0<x-a<\delta\).
Writing the Two Approaches Clearly
The side of approach can be represented by a positive increment \(h\). For a right-hand approach, set \(x=a+h\), where \(h>0\). For a left-hand approach, set \(x=a-h\), also with \(h>0\). Substituting these expressions into the secant slope gives two useful forms:
The negative denominator in the left-hand form matters: \(a-h-a=-h\). Keeping track of it is essential for getting the sign of a left-hand slope correct. The two expressions are the same secant-slope definition written for the two different directions of approach.
Worked Example: A Quadratic at an Interior Point
Worked Example: A Quadratic at an Interior Point
Let \(f(x)=x^2\), and compare secant slopes based at \(a=2\).
First, \(f(2)=2^2=4\). For \(x\ne2\), the secant slope is
The cancellation is valid because \(x\ne2\). For a left-side check, take \(x=1.9\). Then \(f(1.9)=1.9^2=3.61\), so
This agrees with the simplified expression: \(1.9+2=3.9\). For a right-side check, take \(x=2.1\). Then \(f(2.1)=2.1^2=4.41\), so
Again, the simplified expression gives \(2.1+2=4.1\). As \(x\) approaches \(2\) from either side, \(x+2\) approaches \(4\). Thus both one-sided secant-slope limits are \(4\). Substituting the checked inputs into the original quotient reproduces the values obtained from the simplified formula.
Worked Example: Different Limits on the Two Sides
Worked Example: Different Limits on the Two Sides
Let \(f(x)=|x|\), based at \(a=0\). Since \(f(0)=0\), the secant slope is
If \(x<0\), then \(|x|=-x\), so the left-hand secant slope is
For example, at \(x=-0.2\), the function value is \(|-0.2|=0.2\), and the quotient gives \(0.2/(-0.2)=-1\). If \(x>0\), then \(|x|=x\), so the right-hand secant slope is
At \(x=0.2\), the function value is \(0.2\), and the quotient gives \(0.2/0.2=1\). The left-hand slopes remain \(-1\), while the right-hand slopes remain \(1\), however close the inputs get to zero. Therefore the left-hand limit is \(-1\) and the right-hand limit is \(1\); they differ. This example shows why the two directions must be checked separately.
Worked Example: Different Formulas, Matching Limits
Worked Example: Different Formulas, Matching Limits
Consider the piecewise function
At the base input \(a=1\), the first rule gives \(f(1)=1^2=1\). For \(x<1\), the left-hand secant slope is
For instance, at \(x=0.9\), \(f(0.9)=0.9^2=0.81\), and the original quotient gives \((0.81-1)/(0.9-1)=(-0.19)/(-0.1)=1.9\). The simplified expression gives \(0.9+1=1.9\). As \(x\) approaches \(1\) from the left, \(x+1\) approaches \(2\).
For \(x>1\), the second rule gives \(f(x)=2x-1\). The right-hand secant slope is
At \(x=1.1\), \(f(1.1)=2(1.1)-1=1.2\), and the original quotient gives \((1.2-1)/(1.1-1)=0.2/0.1=2\), as does the simplified expression. Thus the right-hand slopes approach \(2\). The formulas on the two sides differ, but both one-sided secant-slope limits equal \(2\). This is a useful reminder that matching behavior does not require identical formulas on either side.
When Do the One-Sided Limits Agree?
Suppose the function is defined for inputs on both sides of \(a\), apart from the excluded base input when forming the quotient. If the left-hand and right-hand limits of \(S_a(x)\) both exist and equal the same finite number \(L\), then the two-sided limit of \(S_a(x)\) exists and equals \(L\). If the two one-sided limits exist but are different, there is no single two-sided limit.
Here is the justification. If both one-sided limits equal \(L\), then for a given \(\varepsilon>0\), the left-hand limit gives a distance \(\delta_L>0\) that keeps \(|S_a(x)-L|<\varepsilon\) on the left. The right-hand limit gives a distance \(\delta_R>0\) with the same property on the right. Choose \(\delta\) to be the smaller of \(\delta_L\) and \(\delta_R\). Whenever \(0<|x-a|<\delta\), \(x\) lies on one side or the other, and the corresponding bound proves \(|S_a(x)-L|<\varepsilon\). This is exactly the definition of the two-sided limit.
Conversely, if the two-sided limit is \(L\), its defining bound holds for every \(x\) close enough to \(a\), whether \(x<a\) or \(x>a\). Restricting that same bound to \(x<a\) proves the left-hand limit is \(L\); restricting it to \(x>a\) proves the right-hand limit is \(L\). Therefore both one-sided limits must agree. In particular, if they have different finite values, no single two-sided limit can satisfy both sides.
A Reliable Way to Compare the Slopes
Write \(a\) and evaluate \(f(a)\), since that same point is used in both secants.
Use \(S_a(x)=\frac{f(x)-f(a)}{x-a}\), with \(x\ne a\).
Use the rule for \(x<a\) on the left and the rule for \(x>a\) on the right.
Factor or simplify the quotient only while keeping the restriction \(x\ne a\) in view.
Determine what each side approaches, then state whether the two results match.
If the function's domain has no inputs on one side of \(a\), then secants from that side cannot be formed. The two-sided comparison requires available inputs arbitrarily close to \(a\) on both sides. Likewise, the base value \(f(a)\) matters: every secant in this comparison connects to the point \((a,f(a))\).
Common Pitfall: Losing a Negative Sign
For \(f(x)=x^2\) based at \(a=2\), a sign error can occur when evaluating the left-side quotient at \(x=1.9\). It is incorrect to write \(1.9^2-4=0.39\). In fact, \(1.9^2=3.61\), so the numerator is \(3.61-4=-0.39\). The denominator is \(1.9-2=-0.1\). Therefore the correct calculation is
not \(0.39/(-0.1)\). The two negative signs produce a positive slope. A good check is to compare the original quotient with the correctly simplified expression \(x+2\): at \(x=1.9\), that expression is also \(3.9\). Also avoid substituting \(x=a\) into the original quotient. Its denominator is zero there; simplification may describe nearby inputs, but it does not make the original quotient defined at the base input.
Why the Comparison Matters
Left- and right-hand secant approaches show whether nearby average rates behave consistently around a chosen input. The quadratic example had matching limits, the absolute-value example had different limits, and the piecewise example had matching limits despite using different formulas on either side. These comparisons are useful when examining graphs, models, or data near a boundary or a change in a rule.
This tutorial concerns the limiting behavior of secant slopes. Agreement between the two sides is a specific observation about that behavior; it does not by itself supply every interpretation or geometric conclusion one might want. The next step in the study of rates of change builds on these one-sided comparisons.
Check Your Understanding
For each question, distinguish carefully between inputs to the left and right of the base point.
- For \(f(x)=x^2\) based at \(a=3\), simplify the secant slope and find its left- and right-hand limits.
- For \(f(x)=|x|\) based at \(a=0\), what are the left- and right-hand secant-slope limits, and do they agree?
- Why must the denominator in the left-hand form using \(x=a-h\) be \(-h\), when \(h>0\)?
- Can two different formulas on either side of a point produce matching one-sided secant-slope limits? Explain using the piecewise example.
- What does a difference between the two one-sided limits tell you about the two-sided limit?
