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Chapter 2: Problem 28

Evaluating limits analytically Evaluate the following limits or state that they do not exist. a. \(\lim _{x \rightarrow-2^{+}} \frac{x^{3}-5 x^{2}+6 x}{x^{4}-4 x^{2}}\) b. \(\lim _{x \rightarrow-2^{-}} \frac{x^{3}-5 x^{2}+6 x}{x^{4}-4 x^{2}}\) c. \(\lim _{x \rightarrow-2} \frac{x^{3}-5 x^{2}+6 x}{x^{4}-4 x^{2}}\) d. \(\lim _{x \rightarrow 2} \frac{x^{3}-5 x^{2}+6 x}{x^{4}-4 x^{2}}\)

Short Answer

Expert verified
Based on the given rational function \(\frac{x^{3}-5 x^{2}+6 x}{x^{4}-4 x^{2}}\), the limit when x approaches -2 does not exist, while the limit when x approaches 2 is equal to -\frac{1}{8}.

Step by step solution

01

Factor the function

Factor both the numerator and the denominator: Numerator: \(x^{3}-5x^{2}+6x = x(x^{2}-5x+6)=x(x-2)(x-3)\) Denominator: \(x^4-4x^2 = x^2(x^2-4)=x^2(x-2)(x+2)\) Thus, the simplified function is: \(f(x)=\frac{x(x-2)(x-3)}{x^2(x-2)(x+2)}\)
02

Evaluate the limits

a. \(\lim _{x \rightarrow-2^{+}} \frac{x(x-2)(x-3)}{x^2(x-2)(x+2)}\) Since the (x-2) term in the numerator and denominator cancels out, the function simplifies to: \(\lim _{x \rightarrow-2^{+}} \frac{x(x-3)}{x^2(x+2)}\) Now, as x approaches -2 from the right side, the denominator goes to 0. Thus, the limit does not exist. b. \(\lim _{x \rightarrow-2^{-}} \frac{x(x-2)(x-3)}{x^2(x-2)(x+2)}\) Similarly, for the left side, the limit does not exist, since the denominator goes to 0. c. \(\lim _{x \rightarrow-2} \frac{x(x-2)(x-3)}{x^2(x-2)(x+2)}\) Since neither the left nor the right limit exists at x = -2, the overall limit does not exist at x = -2. d. \(\lim _{x \rightarrow 2} \frac{x(x-2)(x-3)}{x^2(x-2)(x+2)}\) For x approaching 2, we can simplify the function to: \(\lim _{x \rightarrow 2} \frac{x(x-3)}{x^2(x+2)}\) Now, substitute x = 2: \(f(2) = \frac{2(2-3)}{2^2(2+2)} = \frac{-2}{16} = -\frac{1}{8}\) Thus, the limit of the function at x = 2 is: \(\lim _{x \rightarrow 2} \frac{x^{3}-5 x^{2}+6 x}{x^{4}-4 x^{2}} = -\frac{1}{8}\)

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Key Concepts

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

Analytical Methods in Limit Evaluation
Evaluating limits analytically involves determining the value that a function approaches as the input approaches a certain point. This method is grounded in the algebraic manipulation of the function, leveraging mathematical rules and formulas to simplify complex expressions.

When tackling such problems, particularly in calculus, understanding the nature of the function through its algebraic form is crucial. The practice often includes factoring, rationalizing, or finding a common denominator to break down the expression into manageable parts. Each approach aims to reveal hidden structures within the expression, enabling the determination of the limit more easily.

The key takeaway from analytical methods is to manipulate the expression so that the limit can be evaluated more directly without encountering forms like \(\frac{0}{0}\) or infinity, which require special handling. Utilizing these techniques effectively can provide insights into the behavior of functions near points of interest.
Factoring Polynomials for Simplicity
Factoring polynomials is a fundamental skill that simplifies expressions, particularly in the evaluation of limits. The process involves breaking down a polynomial into simpler, multiplicative components, making it easier to identify and cancel out terms.

For example, when given the expression \(x^3 - 5x^2 + 6x\), factoring yields \(x(x-2)(x-3)\). This transformation not only simplifies the expression but also reveals potential roots or points where the function might have discontinuities.

In the given example, the denominator \(x^4 - 4x^2\) is also factored into \(x^2(x-2)(x+2)\). Identifying common factors in the numerator and denominator allows one to cancel them out, reducing the complexity of the expression. This step is crucial in unraveling indeterminate forms and evaluating limits accurately. Without factoring, you might overlook opportunities to simplify, which could lead to incorrect evaluations.
Understanding One-Sided Limits
One-sided limits are values that a function approaches as the input approaches a specific point from one side鈥攅ither from the left or the right. Evaluating these limits provides insight into the function's behavior around specific points, particularly where discontinuities may occur.

For instance, one-sided limits are denoted as \(\lim_{x \to c^+} f(x)\) and \(\lim_{x \to c^-} f(x)\), representing the right-hand and left-hand limits at a point \(c\), respectively.
  • At \(x \to -2^+\), the function under consideration approaches zero in the denominator from the right, indicating a point where the limit does not exist.
  • Conversely, at \(x \to -2^-\), the approach from the left also results in a denominator approaching zero, showing non-existence of the limit.
Understanding and computing these limits are essential in determining the continuity and behavior of functions across intervals, as they can indicate the presence of jumps or asymptotes.
Exploring Discontinuities
Discontinuities occur in functions where they are not continuous at certain points. These points are crucial because they influence the behavior of limits and can result in non-existent limits.

In the given function, the expression \(\frac{x(x-2)(x-3)}{x^2(x-2)(x+2)}\) exhibits discontinuities at \(x = 2\), \(x = -2\), and \(x = 0\). The common factor \(x-2\) in the numerator and denominator cancels out, indicating a removable discontinuity at \(x = 2\), where the function can still potentially be defined by the reduced expression.

Discontinuities can be of various types such as removable, jump, and infinite, each affecting limits differently. Recognizing and analyzing these discontinuities help in accurately predicting the limit behavior, determining whether the function can be "fixed" at a discontinuity, or if the limit around that point truly does not exist.

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Most popular questions from this chapter

Use the following definitions. Assume fexists for all \(x\) near a with \(x>\) a. We say that the limit of \(f(x)\) as \(x\) approaches a from the right of a is \(L\) and write \(\lim _{x \rightarrow a^{+}} f(x)=L,\) if for any \(\varepsilon>0\) there exists \(\delta>0\) such that $$ |f(x)-L|<\varepsilon \quad \text { whenever } \quad 00\) there exists \(\delta>0\) such that $$ |f(x)-L|<\varepsilon \quad \text { whenever } \quad 0

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