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Use the limit laws to evaluate lim x 6 ( 2 x 1 ) x + 4 . In each step, indicate the limit law applied.

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Limits of polynomial and rational functions

By now you have probably noticed that, in each of the previous examples, it has been the case that lim x a f ( x ) = f ( a ) . This is not always true, but it does hold for all polynomials for any choice of a and for all rational functions at all values of a for which the rational function is defined.

Limits of polynomial and rational functions

Let p ( x ) and q ( x ) be polynomial functions. Let a be a real number. Then,

lim x a p ( x ) = p ( a )
lim x a p ( x ) q ( x ) = p ( a ) q ( a ) when q ( a ) 0 .

To see that this theorem holds, consider the polynomial p ( x ) = c n x n + c n 1 x n 1 + + c 1 x + c 0 . By applying the sum, constant multiple, and power laws, we end up with

lim x a p ( x ) = lim x a ( c n x n + c n 1 x n 1 + + c 1 x + c 0 ) = c n ( lim x a x ) n + c n 1 ( lim x a x ) n 1 + + c 1 ( lim x a x ) + lim x a c 0 = c n a n + c n 1 a n 1 + + c 1 a + c 0 = p ( a ) .

It now follows from the quotient law that if p ( x ) and q ( x ) are polynomials for which q ( a ) 0 , then

lim x a p ( x ) q ( x ) = p ( a ) q ( a ) .

[link] applies this result.

Evaluating a limit of a rational function

Evaluate the lim x 3 2 x 2 3 x + 1 5 x + 4 .

Since 3 is in the domain of the rational function f ( x ) = 2 x 2 3 x + 1 5 x + 4 , we can calculate the limit by substituting 3 for x into the function. Thus,

lim x 3 2 x 2 3 x + 1 5 x + 4 = 10 19 .
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Evaluate lim x −2 ( 3 x 3 2 x + 7 ) .

−13;

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Additional limit evaluation techniques

As we have seen, we may evaluate easily the limits of polynomials and limits of some (but not all) rational functions by direct substitution. However, as we saw in the introductory section on limits, it is certainly possible for lim x a f ( x ) to exist when f ( a ) is undefined. The following observation allows us to evaluate many limits of this type:

If for all x a , f ( x ) = g ( x ) over some open interval containing a , then lim x a f ( x ) = lim x a g ( x ) .

To understand this idea better, consider the limit lim x 1 x 2 1 x 1 .

The function

f ( x ) = x 2 1 x 1 = ( x 1 ) ( x + 1 ) x 1

and the function g ( x ) = x + 1 are identical for all values of x 1 . The graphs of these two functions are shown in [link] .

Two graphs side by side. The first is a graph of g(x) = x + 1, a linear function with y intercept at (0,1) and x intercept at (-1,0). The second is a graph of f(x) = (x^2 – 1) / (x – 1). This graph is identical to the first for all x not equal to 1, as there is an open circle at (1,2) in the second graph.
The graphs of f ( x ) and g ( x ) are identical for all x 1 . Their limits at 1 are equal.

We see that

lim x 1 x 2 1 x 1 = lim x 1 ( x 1 ) ( x + 1 ) x 1 = lim x 1 ( x + 1 ) = 2 .

The limit has the form lim x a f ( x ) g ( x ) , where lim x a f ( x ) = 0 and lim x a g ( x ) = 0 . (In this case, we say that f ( x ) / g ( x ) has the indeterminate form 0 / 0 .) The following Problem-Solving Strategy provides a general outline for evaluating limits of this type.

Problem-solving strategy: calculating a limit when f ( x ) / g ( x ) Has the indeterminate form 0/0

  1. First, we need to make sure that our function has the appropriate form and cannot be evaluated immediately using the limit laws.
  2. We then need to find a function that is equal to h ( x ) = f ( x ) / g ( x ) for all x a over some interval containing a . To do this, we may need to try one or more of the following steps:
    1. If f ( x ) and g ( x ) are polynomials, we should factor each function and cancel out any common factors.
    2. If the numerator or denominator contains a difference involving a square root, we should try multiplying the numerator and denominator by the conjugate of the expression involving the square root.
    3. If f ( x ) / g ( x ) is a complex fraction, we begin by simplifying it.
  3. Last, we apply the limit laws.

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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