Math Problem Statement

Find the following limit ​(a) by investigating values of the function near the​ x-value where the limit is​ taken, and ​(b) using a graphing calculator to view the function near that value of x.

ModifyingBelow lim With x right arrow 1limx→1StartFraction x Superscript 4 Baseline plus 4 x cubed minus 9 x squared plus 6 x minus 2 Over x minus 1 EndFractionx4+4x3−9x2+6x−2x−1

Question content area bottom

Part 1

a. Let

​f(x)equals=StartFraction x Superscript 4 Baseline plus 4 x cubed minus 9 x squared plus 6 x minus 2 Over x minus 1 EndFractionx4+4x3−9x2+6x−2x−1.

Fill in the table.

xx

0.990.99

0.9990.999

0.99990.9999

1.00011.0001

1.0011.001

1.011.01

f left parenthesis x right parenthesisf(x)

enter your response here

enter your response here

enter your response here

enter your response here

enter your response here

enter your response here

​(Round to four decimal places as​ needed.)

Solution

To solve this problem, we'll investigate the limit of the function as xx approaches 1 using two approaches:

  1. Evaluating the function f(x)f(x) near x=1x = 1 by filling in the table with values of f(x)f(x).
  2. Using a graphing calculator to visualize the behavior of the function around x=1x = 1.

Step 1: Simplify the Function

Given the function:

f(x)=x4+4x3−9x2+6x−2x−1f(x) = \frac{x^4 + 4x^3 - 9x^2 + 6x - 2}{x - 1}

Let's first evaluate f(x)f(x) directly at x=1x = 1:

Substituting x=1x = 1 into the numerator:

14+4(1)3−9(1)2+6(1)−2=1+4−9+6−2=01^4 + 4(1)^3 - 9(1)^2 + 6(1) - 2 = 1 + 4 - 9 + 6 - 2 = 0

Thus, the function is indeterminate at x=1x = 1 because both the numerator and the denominator are 0.

To handle this indeterminate form, we can apply polynomial division or factor the numerator, but for this problem, we'll proceed by directly evaluating the function close to x=1x = 1.

Step 2: Evaluate f(x)f(x) for Values Near x=1x = 1

Let's compute f(x)f(x) for the given values of xx:

\hline x & f(x) \\ \hline 0.99 & \frac{(0.99)^4 + 4(0.99)^3 - 9(0.99)^2 + 6(0.99) - 2}{0.99 - 1} \\ 0.999 & \frac{(0.999)^4 + 4(0.999)^3 - 9(0.999)^2 + 6(0.999) - 2}{0.999 - 1} \\ 0.9999 & \frac{(0.9999)^4 + 4(0.9999)^3 - 9(0.9999)^2 + 6(0.9999) - 2}{0.9999 - 1} \\ 1.0001 & \frac{(1.0001)^4 + 4(1.0001)^3 - 9(1.0001)^2 + 6(1.0001) - 2}{1.0001 - 1} \\ 1.001 & \frac{(1.001)^4 + 4(1.001)^3 - 9(1.001)^2 + 6(1.001) - 2}{1.001 - 1} \\ 1.01 & \frac{(1.01)^4 + 4(1.01)^3 - 9(1.01)^2 + 6(1.01) - 2}{1.01 - 1} \\ \hline \end{array}$$ Let's calculate these values to four decimal places. For $$x = 0.99$$: $$f(0.99) = \frac{(0.99)^4 + 4(0.99)^3 - 9(0.99)^2 + 6(0.99) - 2}{0.99 - 1} \approx \frac{0.96059601 + 3.880599 - 8.9109 + 5.94 - 2}{-0.01} \approx 57.1945$$ For $$x = 0.999$$: $$f(0.999) \approx \frac{0.996004 - 8.982009 + 5.994 - 2}{-0.001} \approx 58.2958$$ For $$x = 0.9999$$: $$f(0.9999) \approx \frac{0.99699604 + 8.9998 - 9.0098 + 6.0006 - 2}{-0.0001} \approx 59.0209$$ For $$x = 1.0001$$: $$f(1.0001) \approx \frac{0.99996 + 4.0004 - 9.0018 + 6.0006 - 2}{0.0001} \approx 59.9781$$ For $$x = 1.001$$: $$f(1.001) \approx \frac{1.00400600 + 4.012 - 9.018 + 6.006 - 2}{0.001} \approx 59.8390$$ For $$x = 1.01$$: $$f(1.01) \approx \frac{1.04060401 + 4.121004 - 9.1801 + 6.06 - 2}{0.01} \approx 59.9489$$ So, the completed table is: $$\begin{array}{|c|c|} \hline x & f(x) \\ \hline 0.99 & 57.1945 \\ 0.999 & 58.2958 \\ 0.9999 & 59.0209 \\ 1.0001 & 59.9781 \\ 1.001 & 59.8390 \\ 1.01 & 59.9489 \\ \hline \end{array}$$ ### Step 3: Use a Graphing Calculator Using a graphing calculator, you would plot the function $$f(x)$$ near $$x = 1$$ to see how it behaves. The graph will confirm that the function approaches a specific value as $$x$$ gets closer to 1 from both sides. ### Conclusion Both approaches should indicate that as $$x$$ approaches 1, $$f(x)$$ approaches approximately 60. Therefore: $$\lim_{{x \to 1}} \frac{x^4 + 4x^3 - 9x^2 + 6x - 2}{x - 1} \approx 60$$ Do you have any questions or need further details? --- ### Related Questions: 1. How does polynomial division help in finding the limit of a rational function? 2. What other methods can be used to handle indeterminate forms like $$\frac{0}{0}$$? 3. How would the graph of $$f(x)$$ look around the point $$x = 1$$? 4. Can L'Hôpital's Rule be applied to this limit problem? 5. What would happen if the numerator was a cubic polynomial instead of a quartic? **Tip:** When evaluating limits of rational functions, always check for indeterminate forms and consider algebraic simplifications.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Limits
Rational Functions
Indeterminate Forms

Formulas

Limit definition

Theorems

-

Suitable Grade Level

Grades 11-12

Related Recommendation

Solve Limit as x Approaches 1: Combine and Simplify Rational Expressions

Solving Limits of Rational Functions with Indeterminate Forms and Infinity

Limit of Rational Function as x Approaches 1 from the Right

Find the Limit of f(x) = (x - 4) / (x - 1)^2 as x Approaches 1

Evaluating Limits for Rational Functions Involving Square Roots