One Limit, Two Ways to Describe the Input
A difference quotient measures an average rate of change over an interval. In the h-form, the interval is described by an increment \(h\) from a fixed input \(a\). In the x-to-a form, the nearby input is named \(x\), and \(x\) approaches \(a\).
These forms may look different, but they describe the same nearby inputs. The connection is the equation \(x=a+h\): starting at \(a\) and moving by \(h\) arrives at \(x\). This tutorial derives that connection and explains why it preserves the limit.
The quotients are defined only when their denominators are nonzero. For the h-form this means \(h\ne 0\); for the x-to-a form it means \(x\ne a\). The derivative of \(f\) at \(a\), when it exists as a finite number, is the limit of either quotient as the corresponding input approaches the excluded value.
This equality is not a claim that the quotients have identical written variables. It says that replacing one variable by the other gives the same values at corresponding nearby inputs, so their limits agree.
The Change of Variable
Set \(x=a+h\). Then \(x-a=h\), and \(a+h=x\). Substituting both identities into the h-form gives the x-to-a form:
The key is that this substitution also matches the approaches in the limits. As \(h\) approaches \(0\), \(a+h\) approaches \(a\). Conversely, as \(x\) approaches \(a\), \(x-a\) approaches \(0\). Excluding \(h=0\) corresponds exactly to excluding \(x=a\).
A neighborhood here means an interval around \(a\). For example, there is some positive distance \(\delta\) such that the relevant inputs are defined whenever their distance from \(a\) is less than \(\delta\). In the limit, the center input itself is excluded from the quotient. This condition ensures that both forms use the same nearby inputs. At an endpoint, or when the domain only provides inputs on one side, the corresponding one-sided limits can be compared instead.
Why the Limits Must Agree
A limit describes the values of an expression at inputs close to a target, not necessarily at the target itself. In the h-form the inputs to \(f\) are \(a+h\). In the x-to-a form they are \(x\). The substitution \(x=a+h\) pairs each permitted nonzero increment with exactly one permitted nearby input, and the quotient values in each pair are equal.
For any \(h\ne 0\), choose \(x=a+h\). Then \(x\ne a\).
The distance of \(x\) from \(a\) is \(|x-a|=|h|\). Thus \(h\) is close to \(0\) exactly when \(x\) is close to \(a\).
Since \(a+h=x\) and \(h=x-a\), the two quotients are equal at the matched inputs.
To make the limit argument explicit, suppose the x-to-a quotient has finite limit \(L\). By the definition of that limit, for every \(\varepsilon>0\) there is a \(\delta>0\) such that
For the h-form, take the same \(\delta\). If \(0<|h|<\delta\), set \(x=a+h\). Then \(0<|x-a|=|h|<\delta\), and the quotient identity gives
Therefore the h-form has limit \(L\). The reverse implication follows by the same argument with \(h=x-a\): whenever \(0<|x-a|<\delta\), the increment \(h=x-a\) satisfies \(0<|h|<\delta\), and the quotients are equal. This proves the equivalence in both directions.
Worked Conversions
Worked Example: Convert an x-to-a limit at \(a=4\)
Suppose \(f\) is defined at \(4\) and nearby, and the x-to-a difference quotient has limit \(-2\). Write the equivalent h-form and verify the conversion.
Start with the given limit:
Set \(x=4+h\). Then \(x-4=(4+h)-4=h\), and \(f(x)=f(4+h)\). The expression becomes
As \(x\to4\), \(h=x-4\to0\). Therefore the equivalent h-form is
Check in the reverse direction: set \(h=x-4\), so \(4+h=x\). The h-form quotient becomes \(\frac{f(x)-f(4)}{x-4}\), and \(h\to0\) means \(x\to4\). The original limit and its value are recovered.
Worked Example: Convert an h-form limit at \(a=-1\)
Suppose the h-form limit at \(a=-1\) is given as \(\frac{3}{2}\). Convert it to x-to-a form, showing how the numerator and denominator change.
The given statement is
Solve \(x=a+h\) for \(h\). Since \(a=-1\), this gives \(x=-1+h\), so \(h=x+1\). Also, \(-1+h=x\). Substituting these identities gives
When \(h\to0\), \(x=-1+h\to-1\). Thus
Check by reversing the substitution: \(h=x+1\), and \(-1+h=-1+(x+1)=x\). The denominator \(h=x+1=x-(-1)\). These equalities reproduce the starting quotient and limit.
Worked Example: Match one-sided approaches
Suppose \(f\) is defined for inputs just to the right of \(5\), and consider \(x\to5^+\). Express the corresponding approach in terms of \(h\).
For inputs to the right of \(5\), \(x>5\). Set \(h=x-5\). Then \(h>0\), and as \(x\) approaches \(5\) from the right, \(h\) approaches \(0\) through positive values. The quotient transforms as follows:
For example, \(5<x<5.1\) is equivalent to \(0<h<0.1\): subtract \(5\) throughout the first inequality to obtain the second, and add \(5\) to reverse it. Hence the one-sided limits correspond:
Check the direction: \(h>0\) implies \(5+h>5\), so the function input lies to the right of \(5\), as required. A left-hand approach similarly corresponds to \(h<0\).
A Common Substitution Error
A frequent mistake is to substitute \(x=a+h\) in the numerator but leave the denominator as \(x\), or to replace the denominator \(x-a\) by \(a+h\). The denominator must be transformed using the same substitution as the numerator.
For example, if \(a=3\) and \(x=3+h\), then
not \(3+h\). A mistaken conversion would turn \(\frac{f(x)-f(3)}{x-3}\) into \(\frac{f(3+h)-f(3)}{3+h}\). The correction is
Another error is to set \(h=0\) or \(x=a\) before taking the limit. At those values the denominator is zero, so the quotient is not defined. The limit instead examines nonzero increments, or inputs different from \(a\), that get arbitrarily close to the target.
Why Both Forms Are Useful
The forms emphasize different descriptions of the same change. The h-form highlights the size and direction of an increment from \(a\). The x-to-a form highlights a nearby input approaching \(a\). Choosing either form is a matter of clarity and convenience; neither changes the derivative when the required limit exists.
Check Your Understanding
For each question, focus on how the substitution changes both the quotient and the approach.
- Starting with the h-form at \(a=2\), what substitution gives the x-to-a form, and how does the denominator transform?
- If \(x\to a^-\), what sign do the corresponding increments \(h=x-a\) have?
- Why does \(h\to0\) correspond to \(x\to a\) when \(x=a+h\)?
- In a conversion at \(a=6\), explain why \(x-6\) becomes \(h\), not \(6+h\).
- Why must the quotient be considered at nonzero \(h\) or at \(x\ne a\) before taking its limit?
