Informal Definition 
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, 2x^{2} = 2c^{2}.
1. Sum Rule:  [f_{1}(x) + f_{2}(x)] = f_{1}(x) + f_{2}(x) 
2. Difference Rule:  [f_{1}(x)  f_{2}(x)] = f_{1}(x)  f_{2}(x) 
3. Product Rule:  [f_{1}(x) • f_{2}(x)] = f_{1}(x) • f_{2}(x) 
4. Constant Multiple Rule:  [k • f_{2}(x)] = k • f_{2}(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.
If
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) = a_{n}x^{n} + a_{n1}x^{n1} + . . . + a_{0} 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) x^{2}(2  x) = (2x^{2}  x^{3}) = 2(3)^{2}  (3)^{3} = 18  27 = 9.
b) Same limit, found another way:
x^{2}(2  x) = x^{2}•(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.
Find
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
(x^{3}  8)/(x^{2}  4), so we're not dividing by zero in disguise when we cancel the two (x2)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 (x5)'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 greatestinteger function f(x) = has different righthand and lefthand limits at each integer. As we can see in Figure 2,
Figure 2: At each integer, the greatest integer function y = has different righthand and lefthand 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 Onesided and Twosided Limits 
A function f(x) has a limit as x approaches c if and only if the righthand and lefthand 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 greatestinteger function f(x) = has no limit as x approaches 3. As we saw in Example 13, while . Since the righthand and lefthand 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 righthand and lefthand 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 righthand limit nor a lethand limit as x approaches 0.
Figure 4: The function f = sin(1/x) has neither a righthand nor a lefthand 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. 
Figure 6 
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 (,).
Figure 7
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 x^{2}  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 onesided limits when x = 2, we see that
Figure 8
and
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.
Solution:
 The function f(x) = (1  x^{2})^{1/2} is continuous on the closed interval [1,1].
 The function f(x) = 1/(1  x^{2})^{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) = ¦x^{2}  1¦ is continuous on the entire real line, that is, the interval (,).
Figure 10 
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 11 
Figure 12 
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 xaxis, 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 
