No history yet

Limit Definition

From Average to Instantaneous

The rate of change between two points is straightforward. It's the slope of the line connecting them: the change in the vertical value divided by the change in the horizontal value. For a function f(x)f(x), the average rate of change between a point xx and another point a little bit further along, x+hx+h, is calculated with the line connecting (x,f(x))(x, f(x)) and (x+h,f(x+h))(x+h, f(x+h)).

Average Rate of Change=ΔyΔx=f(x+h)−f(x)(x+h)−x=f(x+h)−f(x)h\text{Average Rate of Change} = \frac{\Delta y}{\Delta x} = \frac{f(x+h) - f(x)}{(x+h) - x} = \frac{f(x+h) - f(x)}{h}

This line is called a secant line. It gives us an average speed over an interval. But what if we want the exact speed at a single moment? This requires finding the instantaneous rate of change. To do that, we need to make the interval between our two points infinitesimally small. We shrink the distance hh until it's practically zero.

As the distance hh approaches zero, the secant line pivots to become a tangent line, a line that touches the curve at exactly one point. The slope of this tangent line is the instantaneous rate of change at that point. This concept is captured formally using a limit—a foundational tool in calculus.

The Limit Definition of a Derivative

We can't just plug h=0h=0 into the difference quotient, as that would lead to division by zero, an indeterminate form. Instead, we use a limit to find the value the expression approaches as hh gets arbitrarily close to zero.

For any function f, the derivative f' of f is given by the equation f' of x equals lim over h right arrow zero of f times left parenthesis x plus h right parenthesis minus f of x divided by h for each value of x in the domain of f for which this limit exists.

This gives us the formal definition of the derivative.

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

The derivative, f′(x)f'(x), is itself a function. It takes an input xx and returns the slope of the tangent line to the original function f(x)f(x) at that specific value of xx. Let's see this in action.

Derivatives from First Principles

Using the limit definition to find a derivative is called finding it from or by definition. It relies on algebraic simplification to resolve the 0/00/0 indeterminate form. Let's start with a simple linear function: f(x)=3x+2f(x) = 3x + 2. We know the slope is always 3, so our result should be f′(x)=3f'(x) = 3.

  1. Substitute into the formula: Replace f(x+h)f(x+h) and f(x)f(x) in the limit definition.
  2. Expand and simplify: Perform the algebra to cancel terms.
  3. Factor out and cancel h: The goal is to eliminate the hh in the denominator.
  4. Evaluate the limit: Once hh is gone from the denominator, you can substitute h=0h=0 to find the result.
f′(x)=lim⁡h→0(3(x+h)+2)−(3x+2)h=lim⁡h→03x+3h+2−3x−2h=lim⁡h→03hh=lim⁡h→03=3\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{(3(x+h) + 2) - (3x + 2)}{h} \\ &= \lim_{h \to 0} \frac{3x + 3h + 2 - 3x - 2}{h} \\ &= \lim_{h \to 0} \frac{3h}{h} \\ &= \lim_{h \to 0} 3 \\ &= 3 \end{aligned}

Now for a more interesting case: a quadratic function, f(x)=x2f(x) = x^2. Here, the slope is always changing, so the derivative should be a function of xx.

f′(x)=lim⁡h→0(x+h)2−x2h=lim⁡h→0(x2+2xh+h2)−x2h=lim⁡h→02xh+h2h=lim⁡h→0h(2x+h)h=lim⁡h→0(2x+h)=2x+0=2x\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} \\ &= \lim_{h \to 0} \frac{(x^2 + 2xh + h^2) - x^2}{h} \\ &= \lim_{h \to 0} \frac{2xh + h^2}{h} \\ &= \lim_{h \to 0} \frac{h(2x + h)}{h} \\ &= \lim_{h \to 0} (2x + h) \\ &= 2x + 0 = 2x \end{aligned}

This result, f′(x)=2xf'(x) = 2x, is a new function that tells us the slope of the tangent line to the parabola y=x2y = x^2 for any value of xx. At x=1x=1, the slope is 2(1)=22(1)=2. At x=−3x=-3, the slope is 2(−3)=−62(-3)=-6. This process of using limits to find the function for the instantaneous rate of change is the core operation of differential calculus.

Ready to test your understanding?

Quiz Questions 1/6

The slope of the secant line connecting two points, (x,f(x))(x, f(x)) and (x+h,f(x+h))(x+h, f(x+h)), on the graph of a function represents the:

Quiz Questions 2/6

To find the instantaneous rate of change, the secant line connecting points (x,f(x))(x, f(x)) and (x+h,f(x+h))(x+h, f(x+h)) is transformed into a tangent line at xx. How is this transformation achieved?

While deriving from first principles is fundamental, you'll soon learn rules that make finding derivatives much faster. However, understanding this limit-based foundation is crucial for grasping why those rules work.