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Chapter 2: Problem 4

Complete the tables and use them to find the given limits. Round calculations to three decimal places. A graphing calculator with a TABLE feature will be very helpful. a. \(\lim _{x \rightarrow 1^{-}}\left(\frac{x^{4}-1}{x-1}\right)\) b. \(\lim _{x \rightarrow 1^{-}}\left(\frac{x^{4}-1}{x-1}\right)\) c. \(\lim _{x \rightarrow 1}\left(\frac{x^{4}-1}{x-1}\right)\)

Short Answer

Expert verified
The limits are 4 for both cases.

Step by step solution

01

Simplifying the Function

First, simplify the function \( f(x) = \frac{x^4 - 1}{x - 1} \). Notice that the numerator can be factored using the difference of squares:\[x^4 - 1 = (x^2 - 1)(x^2 + 1)\]We know \(x^2 - 1\) can be further factored as \((x-1)(x+1)\). So the fraction becomes:\[\frac{(x-1)(x+1)(x^2+1)}{x-1}\]We can cancel \(x-1\) from the numerator and the denominator (for \(x eq 1\)), and the expression simplifies to:\[f(x) = (x+1)(x^2+1)\]
02

Finding the Limit as x Approaches 1 from the Left

To find the limit \( \lim_{x \to 1^-} \frac{x^4 - 1}{x - 1} \), use the simplified function \((x+1)(x^2+1)\).Substitute \(x = 1\) from the left into the simplified function:\[(x+1)(x^2+1) = (1+1)((1)^2+1) = 2 \cdot 2 = 4\]Therefore, the limit is 4.
03

Verifying with a Table for Left-Hand Limit

Construct a table for values of \(x\) approaching 1 from the left (e.g., \(0.9, 0.99, 0.999\)). Compute \(f(x)\) for these values using the simplified expression:- \(x = 0.9\), \(f(0.9) = (0.9+1)((0.9)^2+1) = 3.61\)- \(x = 0.99\), \(f(0.99) = (0.99+1)((0.99)^2+1) = 3.9601\)- \(x = 0.999\), \(f(0.999) = (0.999+1)((0.999)^2+1) = 3.996001\)The table supports the limit result of 4 as the function values approach 4.
04

Finding the Limit as x Approaches 1

For \( \lim_{x \to 1} \frac{x^4 - 1}{x - 1} \), use the same simplified expression:Substitute \(x = 1\) into the simplified function again:\[(x+1)(x^2+1) = (1+1)((1)^2+1) = 2 \cdot 2 = 4\]So the two-sided limit is also 4.
05

Concluding the Solutions

We've confirmed that for both the limit from the left and the general limit at \(x = 1\), the expression simplifies, leading us to the value 4.Thus:- \( \lim_{x \to 1^-}\frac{x^4 - 1}{x - 1} = 4 \)- \( \lim_{x \to 1}\frac{x^4 - 1}{x - 1} = 4 \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference of Squares
The difference of squares is a powerful algebraic technique used to simplify expressions where you have two square terms being subtracted. It's a commonly tested concept in calculus, especially when dealing with limits and rational expressions. When you see an expression like \( x^4 - 1 \), you can break it down using this technique. Here's how it works:
  • The expression \( x^4 - 1 \) can be seen as \((x^2)^2 - 1^2\), aligning it with the difference of squares formula: \(a^2 - b^2 = (a-b)(a+b)\).
  • Apply the formula: \( x^4 - 1 = (x^2 - 1)(x^2 + 1) \).
  • Notice that part of the factor \( x^2 - 1 \) can be further simplified: \( x^2 - 1 = (x-1)(x+1) \).
Thus, \( x^4 - 1 \) fully factors as \( (x-1)(x+1)(x^2+1) \). Recognizing and using this pattern can help simplify complex expressions, especially when you need to find limits. Always try to factor as much as possible to see if terms can cancel, similar to what we did with the \( x-1 \) in this exercise.
Simplifying Rational Expressions
Simplifying rational expressions means reducing a fraction to its simplest form. This is essential in calculus for accurately finding limits and understanding function behavior near points of interest. Here's how to handle the simplification:
  • Start with the expression \( \frac{x^4 - 1}{x-1} \), which is a rational expression with a polynomial numerator and denominator.
  • After factoring the numerator using the difference of squares, you get \( \frac{(x-1)(x+1)(x^2+1)}{x-1} \).
  • Because the term \( x-1 \) appears in both the numerator and denominator, you can cancel it. But be cautious: this simplification holds for \( x eq 1 \).
  • This results in the simplified expression \( (x+1)(x^2+1) \), which is much easier to handle when calculating a limit.
Simplifying before substituting helps ensure you've removed any indeterminate forms, which often appear in limit calculations when the expression had a term that caused division by zero.
Table of Values
Using a table of values is a practical method to estimate the limit of a function as it approaches a certain point, especially if you're unable to directly substitute due to an indeterminate form. When working with limits, here's why and how you create a table:
  • Select values of \( x \) closer and closer to the point of interest, in this case, values approaching 1 from the left such as 0.9, 0.99, and 0.999.
  • Evaluate the simplified function, \( f(x) = (x+1)(x^2+1) \), using these values.
  • The results from each evaluation, such as 3.61, 3.9601, and 3.996001, give you a clearer picture of what the function value is tending towards.
As seen here, the results increasingly get closer to 4 as \( x \) approaches 1 from the left. This consistent approach towards 4 confirms the calculated limit, helping verify your algebraic findings and providing a concrete numerical justification for the limit at that point.

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Most popular questions from this chapter

Suppose that \(L(x)\) is a function such that \(L^{\prime}(x)=\frac{1}{x} .\) Use the Chain Rule to show that the derivative of the composite function \(L(g(x))\) is \(\frac{d}{d x} L(g(x))=\frac{g^{\prime}(x)}{g(x)} .\)

Find the equation for the tangent line to the curve \(y=f(x)\) at the given \(x\) -value. $$ f(x)=(2 x+3)^{3}+(3 x+2)^{2}+4 x-5 \text { at } x=-2 $$

Find \(f^{\prime}(x)\) by using the definition of the derivative. [Hint: See Example 4.] $$ \text { } f(x)=\frac{1}{x^{2}} $$

Find \(f^{\prime}(x)\) by using the definition of the derivative. [Hint: See Example 4.] \(f(x)=\frac{1}{\sqrt{x}}\) [Hint: Multiply the numerator and denominator of the difference quotient by \((\sqrt{x}+\sqrt{x+h})\) and then simplify.]

For each function: a. Find \(f^{\prime}(x)\) using the definition of the derivative. b. Explain, by considering the original function, why the derivative is a constant. $$ f(x)=12 $$
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