|If the values of a function f(x) approach the value L as x approaches c, we say that f(x) has the limit L as x approaches c, and we write|
In simpler terms, as x approaches c, the value of f(x) approaches L.
As the following graph and table suggest, (2x + 1) = 5.
If f is the identity function f(x) = x, then for any value of c:
f(x) = (x) = c
If f is the constant function f(x) = k (the function whose outputs have the constant value k), then for any
value of c:
f(x) = (k) = k
If c is any number, 2x2 = 2c2.
|1. Sum Rule:||[f1(x) + f2(x)]  = f1(x) + f2(x)|
|2. Difference Rule:||[f1(x) - f2(x)]  = f1(x) - f2(x)|
|3. Product Rule:||[f1(x) • f2(x)]  = f1(x) • f2(x)|
|4. Constant Multiple Rule:||[k • f2(x)]  = k • f2(x)|
|5. Quotient Rule:|
We know from Examples 2 and 3 that x = c and k = k. The various properties of limits now let us combine these results to calculate other limits:There is one last property of limits that we must consider. The limit of a function f(x) as x approaches c never depends on what happens when x = c. The limit, if it exists at all, is entirely determined by the values that f has when x is not equal to c. You can see this in the next example.
then f(x) = (x) = 2 while f(2) = 3.
As always, the limit is determined by the function's approach behavior, not by what happens at x = 2 (Figure 1).
Notice that f(x) approaches 2 as x approaches 2 even though f(2) itself is 3.
Figure 1: The graph of .
|Limits of polynomials can always be found by substitution.|
|If f(x) = anxn + an-1xn-1 + . . . + a0 is any polynomial function, then|
|Limits of (many but not all) rational functions can be found by substitution.|
|If f(x) and g(x) are polynomials, then|
a) x2(2 - x) = (2x2 - x3) = 2(3)2 - (3)3 = 18 - 27 = -9.
b) Same limit, found another way:
x2(2 - x) = x2•(2 - x) = (3)2•(2 - 3) = 9•(-1) = -9.
In Example 9, we can use Equation (2) to find the limit of f(x)/g(x) because the value of the denominator, g(x) = x + 2, is not equal to zero when x = 2.
Solution: The denominator is 0 when x = 2, so we cannot calculate the limit by substitution. However, if we factor the numerator and denominator we find that
Is it really all right to cancel terms like this? Yes, because 2 is not in the domain of the original function
(x3 - 8)/(x2 - 4), so we're not dividing by zero in disguise when we cancel the two (x-2)s.
With the (x - 2) out of the way, we can now find the limit by substitution:
(The algebra we just did) (Equation (2) now applies)
|Substitution will not give the limit because x - 5 = 0 when x = 5.|
|We factor the numerator to see if (x - 5) is a factor. It is. We cancel the (x-5)'s, leaving. . .|
|. . .an equivalent form whose limit we can now find by substitution.|
Find the folowing limit:
Solution: By direct substitution we get the indeterminate form 0/0:
In this case, we change the form of the fraction by rationalizing (eliminating the radical in) the numerator:
Therefore, by substitution, we have
The greatest-integer function f(x) = has different right-hand and left-hand limits at each integer. As we can see in Figure 2,
Figure 2: At each integer, the greatest integer function y = has different right-hand and left-hand limits.
The limit of as x approaches an integer n from above is n, while the limit as x approaches n from below is n - 1.
|Relationship between One-sided and Two-sided Limits|
|A function f(x) has a limit as x approaches c if and only if the right-hand and left-hand limits at c exist and are equal. In symbols,|
= L <=> = L and = L (3)
All the following statements about the function y = f(x) graphed in Figure 3 are true.
Figure 3: Example 13 discusses the limit properties of the function y = f(x) graphed here.
At every other point c between 0 and 4, f(x) has a limit as x approaches c.
The greatest-integer function f(x) = has no limit as x approaches 3. As we saw in Example 13, while . Since the right-hand and left-hand limits of f at 3 are not equal, the function has no single limiting value as x approaches 3.
Show that .
Solution: We prove that by showing that the right-hand and left-hand limits are both 0:
Show that the function y = sin(1/x) has no limit as x approaches 0 from either side (Figure 4).
Solution: As x approaches 0, its reciprocal, 1/x, becomes infinite and the values of sin(1/x) cycle repeatedly from -1 to 1. Thus there is no single number L that the function's values all get close to as x approaches zero. This is true even if we restrict x to positive values or to negative values. The function has neither a right-hand limit nor a let-hand limit as x approaches 0.
Figure 4: The function f = sin(1/x) has neither a right-hand nor a left-hand limit as x approaches 0.
|Figure 5: Three points of discontinuity|
|Definition of continuity|
|A function f is said to be continuous at c if the following three conditions are met:|
1. f(c) is defined. 2.f(x) exists. 3. f(x) = f(c)
A function is said to be continuous on an interval (a, b) if it is continuous at each point in the interval.
Determine whether the following functions are continuous on the given interval:
Solution: The graphs of these three functions are shown in Figure 7.
- Since f is a rational function whose denominator is not zero in the interval (0,1), we can conclude that f is continuous on (0,1).
- Since f is undefined at x = 1, we conclude that it is discontinuous at x = 1. Since the denominator is not zero at any other point on the interval, we can conclude that f is continuous at all other points in the interval (0,2).
- Since polynomial functions are defined over the entire real line, we can see that f is continuous on (-,).
|Definition of Continuity on a Closed Interval|
|If f is defined on a closed interval [a, b], continuous on (a, b), and|
the f is said to be(x) continuous on [a, b].
Discuss the continuity of
Solution: We know that the polynomial functions given by 5 - x and x2 - 1 are continuous for all real x. Thus, to conclude that g is continuous on the entire interval [-1,3], we need only check the behavior of g when x = 2. By taking the one-sided limits when x = 2, we see that
Since these two limits are equal, we can conclude that
g(x) = g(2) = 3
Thus, g is continuous at x = 2, and consequently it is continuous on the entire interval [-1,3]. The graph of g is shown in Figure 8.
|Properties of Continuous Functions|
|If b is a real number and f and g are continuous at x = c, then the following functions are also continuous at c.|
|Continuity of a Composite Function|
|If g is continuous at c and f is continuous at g(c), then the composite function given by f°g(x) = f(g(x)) is continuous at c|
Find the intervals for which the three functions shown in Figure 9 are continuous.
- The function f(x) = (1 - x2)1/2 is continuous on the closed interval [-1,1].
- The function f(x) = 1/(1 - x2)1/2 is continuous on the open interval (-1,1). (Note that f is undefined for all x such that ¦x¦>=1.)
- At x = ±1, the limits from the right and left are zero. Thus, the function f(x) = ¦x2 - 1¦ is continuous on the entire real line, that is, the interval (-,).
|Intermediate Value Theorem|
|If f is continuous on [a,b] and k is any number between f(a) and f(b), then there is at least one number c between a and b such that f(c) = k.|
|Figure 13: The graph of y = 1/x|
If f(x) = 3 is the constant function whose outputs have the constant value k, then
f(x) = (3) = 3
f(x) = (3) = 3
1/x = 0
As x approaches , the graph of 1/x gets arbitrarily close to the x-axis, in the following sense: No matter how small a positive number you name, the value of 1/x eventually gets smaller.
We could have multiplied he original fractions together first, to express the product as a single rational function, but it would have taken longer to get the answer that way.
|Summary for Rational Functions|