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Differentiation · Tutorial 391 of 1000

The Derivative as a Limit

Learn how nearby secant slopes lead to a derivative, how to test the limit with sequences, and why differentiability guarantees continuity.

Advanced 9 min read

What You'll Learn

  • Express the slope between two nearby graph points as a difference quotient
  • Rewrite the quotient using an increment from the point of interest
  • Calculate limiting slopes for polynomial and reciprocal examples
  • Detect failure of a derivative limit using sequences from opposite sides
  • Prove the sequential criterion for a limiting slope
  • Prove that differentiability at an interior point implies continuity there

From Secant Slopes to a Limiting Slope

Continuity asks whether function values stay close when input values are close. Differentiation asks a more refined question: as the input moves toward a fixed point, do the slopes of the secant lines settle toward one finite number? That number, when it exists, describes the limiting rate of change at the point.

Let \(f\) be defined on an interval \(I\), and let \(a\) be an interior point of \(I\). For \(x\ne a\), the slope of the line through \((a,f(a))\) and \((x,f(x))\) is

$$ \frac{f(x)-f(a)}{x-a}. $$

As \(x\) approaches \(a\), the horizontal coordinates of those two graph points approach one another. Their heights need not approach one another for an arbitrary function. If the function is differentiable at \(a\), however, the finite limiting slope will also imply continuity there, as we will prove below.

It is often useful to measure the input change by \(h=x-a\). Then \(x=a+h\), and the same secant slope becomes

$$ \frac{f(a+h)-f(a)}{h}, \qquad h\ne0. $$

The value \(h=0\) is excluded because the quotient would have zero in its denominator. The question is whether the quotient approaches a finite number as nonzero \(h\) approaches zero, with \(a+h\) remaining in the domain.

Definition: Let \(f\) be defined on an interval \(I\), and let \(a\) be an interior point of \(I\). If the finite limit $$ \lim_{h\to0}\frac{f(a+h)-f(a)}{h} $$ exists, then \(f\) is differentiable at \(a\). The value of this limit is the derivative of \(f\) at \(a\), denoted \(f'(a)\).

The limit may equivalently be written as \(\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\), since \(h=x-a\). Both forms describe the same secant slopes. The increment form is especially helpful in calculations because expressions such as \(f(a+h)\) can often be expanded and simplified.

Worked Examples: Finding or Rejecting a Limiting Slope

Worked Example: A Cubic at a Nonzero Point

Let \(f(x)=x^3\), and find the limiting slope at \(a=2\). For \(h\ne0\),

$$ \frac{f(2+h)-f(2)}{h} =\frac{(2+h)^3-8}{h}. $$

Expanding the cube gives \((2+h)^3=8+12h+6h^2+h^3\). Substituting this identity into the quotient and cancelling the nonzero factor \(h\),

$$ \frac{(2+h)^3-8}{h} =\frac{12h+6h^2+h^3}{h} =12+6h+h^2. $$

As \(h\to0\), the terms \(6h\) and \(h^2\) tend to zero. Therefore the quotient tends to \(12\), so \(f\) is differentiable at \(2\) and \(f'(2)=12\). The cancellation was performed only for \(h\ne0\); it does not assign a value to the original quotient at \(h=0\).

Worked Example: A Reciprocal at a Positive Input

Let \(f(x)=1/x\), with domain \((0,\infty)\), and consider \(a=2\). For nonzero \(h\) sufficiently close to zero, \(2+h>0\), and

$$ \frac{f(2+h)-f(2)}{h} =\frac{\frac{1}{2+h}-\frac12}{h} =\frac{\frac{2-(2+h)}{2(2+h)}}{h} =-\frac{1}{2(2+h)}. $$

As \(h\to0\), the denominator \(2(2+h)\) tends to \(4\), so the quotient tends to \(-1/4\). Thus \(f\) is differentiable at \(2\), with \(f'(2)=-1/4\). The restriction to \(h\) close enough to zero ensures that the input \(2+h\) remains in the domain.

Worked Example: A Corner with No Limiting Slope

Consider \(f(x)=|x|\) at \(a=0\). Since \(f(0)=0\), its difference quotient is \(|h|/h\) for \(h\ne0\). If \(h>0\), then \(|h|=h\), and the quotient equals \(1\). If \(h<0\), then \(|h|=-h\), and the quotient equals \(-1\).

In particular, along \(h_n=1/n\) the quotients are all \(1\), while along \(k_n=-1/n\) they are all \(-1\). Both sequences of increments tend to zero, but the corresponding quotient values have different limits. Therefore the two-sided limit does not exist, and \(f\) is not differentiable at \(0\).

The function is nevertheless continuous at \(0\), because \(|f(x)-f(0)|=|x|\to0\). This example shows that continuity does not by itself guarantee a limiting slope.

A Sequential Test for the Limiting Slope

The examples suggest a useful way to examine a proposed limit: test the quotient along sequences of nonzero increments that tend to zero. The general sequential criterion for limits gives the following precise statement in this setting.

Theorem (Sequential Criterion for the Derivative Limit): Let \(f\) be defined on an interval \(I\), let \(a\) be an interior point of \(I\), and let \(L\in\mathbb{R}\). Then $$ \lim_{h\to0}\frac{f(a+h)-f(a)}{h}=L $$ if and only if, for every sequence \((h_n)\) of nonzero numbers with \(h_n\to0\), the terms \(a+h_n\) lie in \(I\) eventually and $$ \frac{f(a+h_n)-f(a)}{h_n}\longrightarrow L. $$

Proof. Suppose first that the limit of the quotient is \(L\). Given \(\varepsilon>0\), there is a \(\delta>0\) such that whenever \(0<|h|<\delta\) and \(a+h\in I\), the quotient differs from \(L\) by less than \(\varepsilon\). For any sequence \(h_n\to0\) as in the theorem, eventually \(0<|h_n|<\delta\). Therefore, eventually,

$$ \left|\frac{f(a+h_n)-f(a)}{h_n}-L\right|<\varepsilon. $$

This proves convergence of the quotient sequence to \(L\).

Conversely, suppose the limit of the quotient is not \(L\). By the negation of the epsilon-delta definition of a limit, there is an \(\varepsilon_0>0\) such that for every \(\delta>0\), one can find a nonzero \(h\) with \(a+h\in I\), \(0<|h|<\delta\), and

$$ \left|\frac{f(a+h)-f(a)}{h}-L\right|\geq\varepsilon_0. $$

For each positive integer \(n\), apply this statement with \(\delta=1/n\) and choose such an \(h_n\). Then \(0<|h_n|<1/n\), so \(h_n\to0\), but the quotient values stay at least \(\varepsilon_0\) away from \(L\). They cannot converge to \(L\), contradicting the assumed sequential condition. This proves the converse. \(\square\)

For the absolute-value example, the sequences of positive and negative increments demonstrate the failure directly. More generally, one sequence whose quotient fails to approach a proposed value is enough to disprove that value as the limit. To establish existence, however, the criterion requires the right convergence along every sequence of admissible increments.

Differentiability Implies Continuity

A finite limiting slope imposes more than control over the quotient: it also forces the function values themselves to approach \(f(a)\). The key is to recover the function’s change by multiplying the quotient by the input change.

Theorem (Differentiability Implies Continuity): If \(f\) is differentiable at an interior point \(a\) of an interval \(I\), then \(f\) is continuous at \(a\).

Proof. Define, for nonzero \(h\) sufficiently close to zero,

$$ q(h)=\frac{f(a+h)-f(a)}{h}. $$

By differentiability, \(q(h)\to f'(a)\) as \(h\to0\). In particular, \(q(h)\) is bounded near zero: there are constants \(C\geq0\) and \(\delta_0>0\) such that \(|q(h)|\leq C\) whenever \(0<|h|<\delta_0\) and \(a+h\in I\). This follows directly from the limit definition, for example by requiring \(|q(h)-f'(a)|<1\) and taking \(C=|f'(a)|+1\).

For such \(h\), the definition of \(q(h)\) gives \(f(a+h)-f(a)=h q(h)\). Hence

$$ |f(a+h)-f(a)|=|h|\,|q(h)|\leq C|h|. $$

If \(C=0\), this estimate makes the difference zero for all sufficiently small nonzero \(h\), so continuity follows. If \(C>0\), given \(\varepsilon>0\), choose \(\delta=\min(\delta_0,\varepsilon/C)\). Whenever \(0<|h|<\delta\), the estimate yields \(|f(a+h)-f(a)|\leq C|h|<\varepsilon\). At \(h=0\), the difference is exactly zero. Thus \(f(a+h)\to f(a)\), which is continuity at \(a\). \(\square\)

This proof also clarifies why limiting slopes concern graph points as well as horizontal coordinates. If \(f\) is differentiable at \(a\), continuity ensures that \(f(a+h)\to f(a)\); consequently, the graph points \((a+h,f(a+h))\) approach \((a,f(a))\). Without continuity, nearby horizontal coordinates alone do not imply nearby graph points.

What the Limiting Slope Captures

The difference quotient is an average rate of change over the interval between \(a\) and \(a+h\). A derivative exists when these average rates approach one finite value as the interval shrinks toward \(a\). The limit is not the quotient at \(h=0\), which is undefined; it is the number approached by quotients at nonzero increments.

There are two common pitfalls. First, simplifying a quotient by cancelling \(h\) is legitimate only after recording that \(h\ne0\). Second, a one-sided limiting slope is not automatically a two-sided derivative. At an interior point, increments from both sides are allowed, and their quotient values must approach the same finite number. The absolute-value example fails precisely because its positive and negative increments give different limits.

A practical calculation therefore has three parts: form the quotient using the actual function values, simplify only under the condition \(h\ne0\), and then evaluate the limit as \(h\to0\). When algebra does not settle the issue, the sequential criterion can expose a failure or confirm what must be checked. The next step is to use this limiting-slope idea as a precise definition and develop its consequences systematically.

Check Your Understanding

Use the quotient and limit ideas from this tutorial to answer the following questions.

  1. Why must \(h=0\) be excluded from the difference quotient, even when algebra later cancels a factor of \(h\)?
  2. For \(f(x)=x^3\) at \(a=2\), what expression remains after simplifying the quotient, and what is its limit as \(h\to0\)?
  3. For \(f(x)=|x|\) at \(0\), calculate the quotient separately for \(h>0\) and \(h<0\). What does this show about differentiability there?
  4. In the sequential criterion, why does one sequence of increments whose quotient fails to approach \(L\) rule out \(L\) as the limit?
  5. In the proof that differentiability implies continuity, where is boundedness of the difference quotient used?