We write

lim x→a f(x) = L

or

f(x) → L as x → a

A more precise way of phrasing the definition is that we can make f(x) be as close to L as we like by making x be sufficiently close to a. We write

lim x→a+ f(x) = L

or

f(x) → L as x → a+

lim x→a- f(x) = L

or

f(x) → L as x → a-

We write

lim x→+∞

f(x)

=

L

lim x→-∞

f(x)

=

L

1. As x → 3, the quantity 3x²-4x+2 approaches 17, and hence

lim x→3

(3x2-4x+2)

=

27

2. Notice that the function

x2 - 9 x - 3

x2 - 9 x - 3

=

(x - 3)(x + 3) x - 3

=

x + 3,

x2 - 9 x - 3

→ 6

as

x → 3,


or

lim x→3

x2 - 9 x - 3

=

6

There are many more examples in the on-line tutorial on limits computed algebraically.

• Make a table of values for f(x) using values of x that approach a closely from either side.
• If the limit exists, then the values of f(x) will approach the limit as x approaches a from both sides.
• The more accurately you wish to evaluate this limit, the closer to a you will need to choose the values of x.
• For a limit as x → +∞, use positive values of x getting larger and larger.
• For a limit as x → -∞, use negative values of x getting larger and larger in magnitude.

1. To estimate

lim x→3

x2 - 9 x - 3

, constuct a table with values of x close to 3 on either side:

x approaching 3 from the left →		x approaching 3 from the right ←
x 2.9 2.99 2.999 2.9999 f(x) = x2 - 9 x - 3 5.9 5.99 5.999 5.9999	3	3.0001 3.001 3.01 3.1 6.0001 6.001 6.01 6.1

Since the values of f(x) appear to be approaching 6 as x approaches 3 from either side, we estimate that the limit is 6.

2. To estimate

lim x→ +∞

x2 - x + 1 2x2 - 3

, constuct a table with values of x approaching +∞:

x approaching +∞   →
x 10 100 1000 10,000 f(x) = x2 - x + 1 2x2 - 3 0.461929 0.495124 0.499501 0.49995	+∞

Since the values of f(x) appear to be approaching 0.5 as x approaches 3 from either side, we estimate that the limit is 0.5.

• Draw the graph of f(x) either by hand or using technology, such as a graphing calculator.
• If you want to calculate a limit as x → a for a real number a, position your pencil point (or the graphing calculator "trace" cursor) on a point of the graph to the left of x = a.
• Move the point along the graph toward x = a from the left and read the y-coordinate as you go. The value the y-coordinate approaches (if any) is then the limit

  lim x→a- f(x)

• Repeat the above two steps, but this time starting from a point on the graph to the right of x = a, and approach x = a along the graph from the right. The value the y-coordinate approaches (if any) is then

  lim x→a+ f(x)

• If the left and right limits both exist and have the same value L, then

  lim x→a+ f(x)

  exists and equals L.

• If you want to calculate a limit as x → +∞, position your pencil point (or the graphing calculator "trace" cursor) on a point near the right edge of the graph, and move the pencil along the graph to the right, estimating the y-coordinate as you go. The value the y-coordinate approaches (if any) is then the limit

  lim x→+∞+ f(x)

  For x → -∞, start toward the left edge, and move your pencil toward the left.

1. Check to see whether f is a closed form function. These are functions specified by a single formula involving constants, powers of x, radicals, exponentials and logarithms, combined using arithmetic operations and composition of functions.
   1. If a is in the domain of f, then lim_{x → a} f(x) = f(a).
   2. If a is not in the domain of f, but f(x) can be reduced by simplification to a function with a in its domain, then (a) applies to the reduced form of the function.
   3. If a is not in the domain of f, and you cannot simplify the function as in (b), then simplify as much as possible, and evaluate the limit by the numerical approach, or, if you know its graph, by the graphical approach.
2. If f is not closed-form, and a is a point at which the definition of f changes, compute the left limit and right limit seperately, and check whether they agree.

Limits as x → ±∞

To compute a limit of the form

lim x→±∞

f(x)

algebraically:

If x is approaching ±∞, then check to see whether f(x) is a ratio of polynomials. If it is, then you can ignore all but the highest powers of x in the numerator and denominator. This simpler function will have the same limit as f.

1.

Consider the limit

lim x→1

x2 - 9 x - 3

.

Notice that the function

f(x) =

x2 - 9 x - 3

is closed form, and a = 1 is in its domain.

lim x→1

x2 - 9 x - 3

=

1 - 9 1 - 3

= 4

2.

Now consider

lim x→3

x3 - 9 x - 3

.

lim x→3

x2 - 9 x - 3

=

lim x→3

(x - 3)(x + 3) x - 3

=

lim x→3

x + 3.

lim x→3

x2 - 9 x - 3

=

lim x→3

x + 3

= 3 + 3 = 6.

3.

Consider

lim x→+∞

x3 + x2 - 9 2x3 - x - 3

.

lim x→+∞ x3 + x2 - 9 2x3 - x - 3	=	lim x→+∞ x3 + x2 - 9 2x3 - x - 3
	=	lim x→+∞ x3 2x3
	=	lim x→+∞ 1 2 = 1 2	(Cancel the x2)

Press here for the on-line tutorial on limits.

A function f is continuous at a if lim_{x → a}  f(x) exists, and is equal to f(a).

The function f is said to be continuous on its domain if it is continuous at each point in its domain. The algebraic approach to limits above is based on the fact that any closed form function is continuous on its domain.

The function f(x) = 3x²-4x+2 is a closed form function, and hence continuous at every point in its domain (all real numbers).

The function

g(x)

=

4x2+1 x - 3

On the other hand, the function

h(x)

=

-1 if -4 ≤ x < -1 x if -1 ≤ x ≤ 1 x2-1 if  1 < x ≤ 2

The average rate of change of f(x) over the interval [a, b] is

Average rate of change

=

Δf Δx

=

f(b) - f(a) b - a

.

The average rate of change of f(x) over the interval [a, a+h] is

Average rate of change

=

f(a+h) - f(a) h

.

Units: The units of the average rate of change are units of f per unit of x.

Let f(x) = 2x² - 4x + 1. Then the average change of f(x) over the interval [2, 4] is

Average rate of change	=	f(4) - f(2) 4 - 2
	=	17 - 1 2 = 8.

Interpretation: If, say f(x) represents the annual profit of your company (in millions of dollars) and x represents the year since January 1990, then the units of measurement of the average rate of change are millions of dollars per year. Thus, your company made an average annual profit of $8 million per year over the period January 1992 to January 1994.

Use the handy little utility below to compute the average change of the above function f(x) over other intervals. Enter the x-coordinates (a and b in the formula), leave everything else blank, and press "Compute." (You can also change the function to anything you like, using standard technology formatting.)

The instantaneous rate of change of f(x) at x = a is given by taking the limit of the average rates of change (computed by the difference quotient) as h approaches 0.

Instantaneous rate of change

=

lim h→0

f(a+h) - f(a) h

.

f'(a)

=

lim h→0

f(a+h) - f(a) h

.

Note:

• f'(a) = Instantaneous rate of change of f at the point a.
• f'(x) = Instantaneous rate of change of f at the point x.

Let f(x) = 2x² - 4x + 1, as above. Then the instantaneous change of f(x) at x = 2 is

f'(2) = 4.

Interpretation
If, say f(x) represents the annual profit of your company (in millions of dollars) and x represents the year since January 1990, then the units of measurement of the instantaneous rate of change are millions of dollars per year. Thus, your company's annual profit was increasing at a rate of $4 million per year at the start of 1992.

To compute an approximate value of f'(a) (for a given value of a) numerically, one can use either:

1. A table of values
2. A quick approximation

Using a Table
In a table, compute a succession of values of difference quotients

f(a+h) - f(a) h

A Quick Approximation
Use a single small value of h and compute the difference quotient:

f'(a)

≈

f(a + 0.0001) - f(a) 0.0001

Forward Difference Quotient

Another Quick Approximation: Balanced Difference Quotient
The following formula often gives a better estimate of the derivative

f'(a)

≈

f(a+0.0001) - f(a-0.0001) 0.0002

Balanced Difference Quotient

Derivative Calculator (Balanced Difference Quotient Approximation)
Enter a function, enter the point a, adjust h as you want, and press "Compute".

f(x) = a =     h =

f'(a)

   

Continuing with the example f(x) = 2x² - 4x + 1, let us compute an approximate value of f'(2).

Using a Table: The difference quotient (with a = 2) is

	f(2+h) - f(2) h
=	2(2+h)2-4(2+h)+1 - (2(2)2-4(2)+1) h

As h gets smaller, we see that the value gets closer and closer to 4, so we conclude

f'(2) ≈ 4.

Using A Quick Approximation (Forward Difference Quotient): We use the formula (with a = 2)

f'(2)	≈	f(2+0.0001) - f(2) 0.0001
	=	f(2.0001) - f(2) 0.0001
	=	1.00040002 - 1 0.0001	=	4.0002.

Notice that the "quick approximation" method does not give the exact answer, but the balanced difference quotient will in this case (see opposite).

Secant and Tangent Lines
The slope of the secant line through the points on the graph of f where x = a and x = a+h is given by the slope of the line PQ in the following diagram:

msec = Slope of secant line through P and Q = f(a+h) - f(a) h

This is also the formula for the average rate of change of f over [a,a+h]. So,

The slope of the tangent line through the point on the graph of f where x = a is obtained by moving the point Q closer to P; in other words, by letting h approach 0:

mtan = slope of tangent

=

lim h→0

f(a+h) - f(a) h

=

f'(a)

This is also the formula for the instantaneous rate of change of f at the point a. So,

We can approximate the slope of the tangent through the point where x = a by using the balanced difference quotient,

mtan

≈

f(a+0.0001) - f(a0.0001) 0.0002

.

Zooming In
We can also interpret the derivative, or slope of the tangent, at a given point on the graph as the slope of the (almost) straight line obtained by "zooming-in" to that point on the curve.(See opposite.)

Extra Topic: Graph of the Derivative
Press here for on-line on how to obtain the graph of f' from the graph of f.

Continuing with the example f(x) = 2x² - 4x + 1,

Slope of secant line through points where x = 2 and x = 3
	= Average rate of change of f(x) over [2, 3] = 6   (see calculation above)

Slope of tangent line through point where x = 2
	= Instantaneous rate of change of f(x) at x = 2 = 4   (see calculation above)

Here is the graph with these two lines shown.

Zooming In

Here is an illustration of zooming in to a point on a graph where x = 0.75.

To compute the derivative of a function algebraically, proceed as follows.

1. Write down the definition of the derivative,

   f'(x)

   =

   lim h→0

   f(x+h) - f(x) h

   .

2. Substitute for f(x+h) and f(x)

   You can use an actual value for x if you are asked, say, to compute f'(3), or just leave it as x if you are asked for the derivative function f'(x) .

3. Simplify the numerator in order to factor out an "h." Then cancel the "h"s and take the limit to obtain the answer.

   Sometimes, you need to do more than just simplify the numerator...

Going back to our first example, f(x) = 2x² - 4x + 1, let us now calculate f'(x) algebraically by following the steps in the adjacent window.

f'(x)	=	lim h→0 f(x+h) - f(x) h
	=	lim h→0 (2(x+h)2-4(x+h)+1) - (2x2-4x+1) h
	=	lim h→0 2x2+4xh+2h2-4x-4h+1-2x2+4x-1 h
	=	lim h→0 4xh+2h2-4h h
	=	lim h→0 h(4x+2h-4) h
	=	lim h→0 (4x+2h-4)
	=	4x-4

Thus,  f'(x) = 4x-4.

Go to the tutorial on average rates of change for practice in computing average rates of change algebraically (whatg we did above up to the last step), or to tutorial on computing the derivative algebraically and scroll down to the box called "Computing the Derivative Algebraically." The "Help" button brings up the complete solution.

For an object moving in a straight line with position s(t) at time t, the average velocity from time t to time t+h is given by the difference quotient

vaverage

=

s(t+h) - s(t) h

.

The instantaneous velocity at time t is given by

v(t)

=

s'(t)

=

lim h→0

s(t+h) - s(t) h

.

Suppose the position of a moving object is given by

s(t) = t² -2t+4 miles

s'(t) = 2t-2 miles per hour.

s'(3) = 4 miles per hour.

Power Rule
If f(x) = xⁿ, then f'(x) = nx^n-1. This is true for any real n. In differential notation, this reads

d dx xn

=

nxn - 1

Sum and Constant Multiple Rules
If f'(x) and g'(x) exist, and c is a constant, then

(a) [f(x)   ±   g(x)]' = f'(x)   ±   g'(x)

(B) [cf(x)]' = cf'(x) .

(a)

d dx

[f(x) ± g(x)] =

d dx

[f(x)] ±

d dx

[g(x)]

(B)

d dx

[cf(x)] = c

d dx

[f(x)]

In words:

The derivative of a sum is the sum of the derivatives, and the derivative of a difference is the difference of the derivatives.

The derivative of c times a function is c times the derivative of the function.

d dx x3.2

=

3.2x2.2

d dx 4x - 1

=

- 4x - 2

d dx 3x1.1 + 4x - 2

=

3.3x0.1 - 8x - 3

d dx 1 x	=	d dx x - 1
	=	- x - 2
	=	- 1 x2

Want some practice? Try the interactive tutorial or try some exercises.

If Q(x) represents any quantity such as cost, revenue, profit or loss on the sale of x items, then Q'(x) is called the marginal quantity. Thus, for instance, the marginal cost measures the increase in total cost per item. This is effectively the cost of each additional item.

The marginal cost is distinct from the average cost, which measures the average of the total cost of the first x items. Average cost is given by

C(x)	=	C(x) x
	=	Total Cost Number of Items

Suppose the cost of (the first) x items is given by

C(x) = 4x^0.2 - 0.1x pounds Sterling.

C'(x) = 0.8x^-0.8 - 0.1 pounds Sterling per unit.

The average cost of the first three units is

C(3)	=	C(3) 3
	≈	4.6829 3
	≈	1.56 pound Sterling/item