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8.4: Polynomial Function Limits

Created by: CK-12

You may recall from a prior lesson on "Operations on Functions" that many of the rules you have learned to use when solving basic Algebra equations often apply to the manipulation of entire equations also. You will find in this lesson that a similar concept applies to operations on limits.

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- KhanAcademy: Limit Examples (part 3)

Guidance

In this lesson, you will evaluate the limits of functions from an algebraic perspective. First you will require the proper tools for the job, the theorems necessary to calculate limits. These theorems are outlined in the two boxes below. Note that you need not necessarily memorize the entire list, but you will want to keep them handy for the lessons on rational function limits and computing one-sided limits.

Box #1: Important Theorems of Limits

Let a be a real number and suppose that \lim_{x\rightarrow a}f(x) = L_1 and \lim_{x\rightarrow a}g(x) = L_2 .

Then:

1.  \lim_{x\rightarrow a} [f(x) + g(x)] = \lim_ {x\rightarrow a}f(x) +  \lim_{x\rightarrow a}g(x) = L_1 + L_2 , meaning the limit of the sum is the sum of the limits.

2.  \lim_{x\rightarrow a} [f(x) - g(x)] =  \lim_ {x\rightarrow a}f(x) -  \lim_{x\rightarrow a}g(x) = L_1 - L_2 , meaning the limit of the difference is the difference of the limits

3.  \lim_{x\rightarrow a} [f(x)g(x)] = ( \lim_ {x\rightarrow a}f(x))  ( \lim_{x\rightarrow a}g(x)) = L_1L_2 , meaning the limit of the product is the product of the limits.

4.  \lim_{x\rightarrow a} \frac{f(x)}{g(x)} =\frac{\lim_{x\rightarrow a} f(x)} {\lim_{x\rightarrow a} g(x)} = \frac {L_1} {L_2} L_2 \neq 0 , meaning the limit of a quotient is the quotient of the limits (provided that the denominator does not equal zero.)

5.  \lim_{x\rightarrow a} \sqrt[n]{f(x)}= \sqrt[n] {  \lim_{x\rightarrow a} f(x)} = \sqrt[n] L_1  L_1 > 0 if n is even, meaning the limit of the n th root is the n th root of the limit.

Other useful results follow from the above theorems:

Box #2

1. If a and k are real numbers, then \lim_{x\rightarrow a}k = k . That is, if f ( x ) = k , a constant function, then the values of f ( x ) do not change as x is varied, thus the limit of f ( x ) is k .

2. If a is a real number then \lim_{x\rightarrow a} x= a . That is, since f ( x ) = x is an identity function (its input equals its output), then as x a , f ( x ) = x a .

3. \lim_{x \rightarrow a} (k \cdot f(x)) = (  \lim_{x \rightarrow a} k) \cdot (\lim_{x \rightarrow a} f(x)) = k \cdot ( \lim_{x \rightarrow a} f(x))

4. \lim_{x \rightarrow a} x^n = (  \lim_{x \rightarrow a} x)^n = a^n

Finally, we have one more theorem:

Theorem: The limit of a polynomial

For any polynomial f ( x ) = c n x n + . . . + c 1 x + c 0 and any real number a ,
\lim_{x \rightarrow a}f(x) = c_n(a)^n + . . . + c_1(a) + c_0
\lim_{x \rightarrow a}f(x) = f(a)
In other words, the limit of the polynomial is simply equal to f ( a ).

Example A

Use the theorems above to find \lim_{x \rightarrow 1}(x^2 - 3x + 4) and justify each step.

NOTE:
This particular example was specifically chosen as "A", since you can verify your practice application of the theorems described above by simply substituting x = 1 directly into the polynomial:
\lim_{x \rightarrow 1}(x^2 - 3x + 4) = (1)^2 - 3(1) + 4 = 2

Solution

Using Equation (1) of Box #1 (the limit of the sum is the sum of the limits) we get

\lim_{x \rightarrow 1}(x^2 - 3x + 4) = \lim_{x \rightarrow 1}x^2 + \lim_{x \rightarrow 1}(-3x) + \lim_{x \rightarrow 1}4
From Equation (4) of Box #2, the first term becomes
\lim_{x \rightarrow 1}x^2 = (1)^2 = 1
From Equations (2) and (3) of Box #2, the second term becomes
\lim_{x \rightarrow 1}(-3x) = -3 \lim_{x \rightarrow 1}x = (-3)(1) = -3
Finally, from Equation (1) of Box #2, the third term becomes
\lim_{x \rightarrow 1}4 = 4

Thus the limit of the above polynomial is

\lim_{x \rightarrow 1}(x^2 - 3x + 4) = 1 + (-3) + 4 = 2

Example B

Find \lim_{x\rightarrow 3} (4x^3 - 4x - 5)

Solution

According to the theorem above, \lim_{x\rightarrow 3}(4x^3 - 4x - 5) = 4(3)^3 - 4(3) - 5 =91

Example C

Find \lim_{x\rightarrow 5} \frac{x^2 - 4} {3x^2 - 2}

Solution

Using Equation (4) of Box #1 (the limit of the quotient is the quotient of the limit),

\lim_{x \rightarrow 5} \frac{x^2 - 4} {3x^2 - 2}= \frac {\lim_{x \rightarrow 5} (x^2 - 4)} {\lim_{x \rightarrow 5}(3x^2 - 2)}
Making use of the limit of the polynomials theorem, we obtain,
\lim_{x \rightarrow 5} \frac{x^2 - 4} {3x^2 - 2} = \frac{(5)^2 - 4} {3(5)^2 - 2}
\mathit = \frac{21} {73}

Vocabulary

The limit theorems are a series of statements describing the effects of various mathematical operations on limits.

Guided Practice

Questions

1) Find the limit: \lim_{x \to 256} \sqrt[4]{x}

2) Find the limit: \lim_{x \to -3} \left( \frac{x^2 + 3x}{x + 3}\right)

3) Given:

\lim_{x \to c} f(x) = 7
\lim_{x \to c} g(x) = 3

Find: \lim_{x \to c} f(x) g(x)

4) Given:

f(x) = 4x^2 + 3x + 4
g(x) = 2x^2 -5x -2

Find: \lim_{x \to -2} f(g(x))

5) Given:

\lim_{x \to c} f(x) = -2
\lim_{x \to c} g(x) = 2

Find: \lim_{x \to c} \frac{f(x)}{g(x)}

Solutions

1) Since the function has no specific applicable discontinuities, just solve for x = 256

\therefore  \lim_{x \to 256} \sqrt[4]{x} = 4

2) To find \lim_{x \to -3} \left( \frac{x^2 + 3x}{x + 3}\right)

\left( \frac{x(x + 3)}{x + 3}\right) ..... First factor the numerator
Since both numerator and denominator contain (x + 3) the function output is the same as the input for any number except where (x + 3) was undefined, -3.
Since x = y for any number other than -3, as we get closer and closer to -3 on the input, we get closer and closer to -3 on the output (we can get as close as we like, we just can't hit it!).
\therefore \lim_{x \to -3} \left( \frac{x^2 + 3x}{x + 3}\right) = -3

3) This example uses rule #3 from above: the limit of the product is the product of the limits. Which means that we need to find the limit of each function, then multiply the limits.

\lim_{x \to c} f(x) = 7
\lim_{x \to c} g(x) = 3

Since we are given the limits, we simply multiply 7 \cdot 3 = 21

\therefore \lim_{x \to c} f(x) g(x) = 21

4) Since these are both continuous functions, we can use \lim f(g(x)) = f \left(\lim g(x)\right)

\lim_{x \to -2} 2x^2 -5x -2 = 16
f(16) = 4(16)^2 + 3(16) + 4 = 1076
\therefore \lim_{x \to -2} f(g(x)) = 1076

5) This example uses rule #4 from above: The limit of a quotient is the quotient of the limits (provided that the denominator does not equal zero).

\lim_{x \to c} f(x) = -2
\lim_{x \to c} g(x) = 2
Since we are given the limits of f(x) and g(x) we just need to divide
\frac{-2}{2} = -1
\therefore \lim_{x \to c} \frac{f(x)}{g(x)} = -1

Practice

Find the Limit

  1. \lim_{x \to 2} -5x - 2=
  2. \lim_{x \to \frac{\pi}{6}} csc (x) =
  3. \lim_{x \to 256} \sqrt[4]{x}=
  4. \lim_{x \to0 } \frac{x^2 - 2x}{x}=
  5. \lim_{x \to 1} 4x =

Operations on Limits: Find the Limit

  1. \lim_{x \to -2}\sqrt{-3x + 3} =
  2. Given: \lim_{x \to c} f(x) = -1 and \lim_{x \to c } g(x) =  -1 Solve: \lim_{x \to c} f(x) - g(x) =
  3. Given: f(x) = -2x^2 + x - 3 and g(x) = -x + 2 Solve: \lim_{x \to -5} f(g(x))=
  4. Given: \lim{x \to c} f(x) = -2 and \lim_{x \to c} g(x) = -5 Solve: \lim_{x \to c} f(x) + g(x)=
  5. Given: f(x) = 2x^2 - 2x + 3 and g(x) = -3x^2 - 4x - 5 Solve: \lim_{x \to 0} = g(f(x))=
  6. \lim_{x \to-1} \sqrt{-5x^3 - 5x^2 - 5x - 3} =
  7. Given: \lim_{x \to c} f(x) = 7 and \lim_{x \to c} g(x) = -3 Solve: \lim_{x \to c} f(x) - g(x) =
  8. Given: f(x) = x^2 - 2x + 2 and g(x) = 4x^2 - 2x - 2 Solve: \lim_{x \to-1 }f(g(x)) =
  9. \lim_{x \to 0} \sqrt{-4x^2 + 4x - 3}=
  10. Given \lim_{x \to c} f(x) = 5 and \lim_{x \to c} g(x) = 8 Solve: \lim_{x \to c} f(x) + g(x)=
  11. Given: f(x) = -x^2 -3x -4 and g(x) = 2x^2 - 5x + 2 Solve: \lim_{x \to 4} f(g(x)) =
  12. \lim_{x \to-6 } \sqrt{-5x - 4} =

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Date Created:

Nov 01, 2012

Last Modified:

May 27, 2014

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