Problem 24 - Calculus Early Transcendentals

Question

Determining limits analytically Determine 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. \(\lim _{x \rightarrow 2} \frac{x^{3}-5 x^{2}+6 x}{x^{4}-4 x^{2}}\)

Step-by-Step Solution

Answer

Question: Find the limits of the function \(f(x)=\frac{x^3-5x^2+6x}{x^4-4x^2}\) for the following cases: a. As \(x \rightarrow -2^+\) b. As \(x \rightarrow -2^-\) c. As \(x \rightarrow -2\) d. As \(x \rightarrow 2\) Answer: a. The limit as \(x \rightarrow -2^+\) does not exist. b. The limit as \(x \rightarrow -2^-\) does not exist. c. The limit as \(x \rightarrow -2\) does not exist. d. The limit as \(x \rightarrow 2\) is \(-\frac{1}{8}\).

Step 1: a. Limit as \(x \rightarrow -2^+\)

To find the limit as \(x \rightarrow -2^+\), we first need to simplify the function: \(\frac{x^{3}-5 x^{2}+6 x}{x^{4}-4 x^{2}} = \frac{x(x^2 - 5x + 6)}{x^2(x^2 - 4)}\). Now, factor the numerator and the denominator: \(\frac{x(x-2)(x-3)}{x^2(x+2)(x-2)}\). Then, cancel the common factors: \(\frac{x(x-3)}{x^2(x+2)}\). Now, apply the limit as \(x \rightarrow -2^+\): \(\lim_{x \rightarrow -2^+} \frac{x(x-3)}{x^2(x+2)} = \frac{-2(-5)}{(-2)^2(0^+)} = \frac{10}{0^+}\). Since the denominator approaches zero, but the numerator is non-zero, the limit does not exist. So, we state that \(\lim _{x \rightarrow-2^{+}} \frac{x^{3}-5 x^{2}+6 x}{x^{4}-4 x^{2}}\) does not exist.

Step 2: b. Limit as \(x \rightarrow -2^-\)

Using the simplified expression from part a, apply the limit as \(x \rightarrow -2^-\): \(\lim_{x \rightarrow -2^-} \frac{x(x-3)}{x^2(x+2)} = \frac{-2(-5)}{(-2)^2(0^-)} = \frac{10}{0^-}\). Since the denominator approaches zero, but the numerator is non-zero, the limit does not exist. So, we state that \(\lim _{x \rightarrow-2^{-}} \frac{x^{3}-5 x^{2}+6 x}{x^{4}-4 x^{2}}\) does not exist.

Step 3: c. Limit as \(x \rightarrow -2\)

Since both one-sided limits as \(x \rightarrow -2^+\) and \(x \rightarrow -2^-\) do not exist, the general limit as \(x \rightarrow -2\) also does not exist. So, we state that \(\lim _{x \rightarrow-2} \frac{x^{3}-5 x^{2}+6 x}{x^{4}-4 x^{2}}\) does not exist.

Step 4: d. Limit as \(x \rightarrow 2\)

Using the simplified expression from part a, apply the limit as \(x \rightarrow 2\): \(\lim_{x \rightarrow 2} \frac{x(x-3)}{x^2(x+2)} = \frac{2(-1)}{(2)^2(4)} = \frac{-2}{16}\). Thus, \(\lim _{x \rightarrow 2} \frac{x^{3}-5 x^{2}+6 x}{x^{4}-4 x^{2}} = -\frac{1}{8}\).

Key Concepts

Limit Existence

The concept of whether a limit exists is important in calculus. We determine if a limit exists by analyzing the behavior of the function as it approaches a particular point. If both sides of the limit approach the same value, the limit exists; otherwise, it does not.

• If a limit approaches infinity or negative infinity, it is considered not to exist, as the function does not tend towards a finite value.
• A limit may also be nonexistent if one-sided approaches from the left and right do not coincide.

In our exercise, we discovered that the limit for parts (a), (b), and (c) does not exist due to the one-sided limits not matching each other and both differing at an infinite approach. Always check both left-sided and right-sided tendencies to conclude limit existence.

One-Sided Limits

One-sided limits analyze the behavior of a function as it approaches a specific value only from one side—either from the left or the right.

• Approaching from the left is denoted as \( x \rightarrow c^- \) and from the right as \( x \rightarrow c^+ \).
• These help in understanding functions with jumps or infinite discontinuities at particular points.

In the solution, when analyzing the limit as \( x \rightarrow -2^+ \) and \( x \rightarrow -2^- \), we noticed that the function approaches different infinities based on the numerator's calculation. This insight about differing infinite tendencies confirms that the overall limit does not exist there. This examination of one-sided limits is critical for understanding function behavior near specific points.

Limit Evaluation

Limit evaluation involves calculating the behavior of functions as they tend toward a particular input value. This can be done analytically by substituting and simplifying expressions. When evaluating limits, particularly with rational functions:

• In some cases, direct substitution can solve the limit; however, when a function results in an indeterminate form like \( \frac{0}{0} \), more advanced methods are required.
• Techniques such as simplifying factored forms or rationalizing might be necessary.
• If the simplified function leads to a non-zero over zero form, the limit leads to infinity, indicating that the limit does not exist.

For instance, as evaluated in step (d), the correct factorization and simplifying led us to find a numerical solution of \( -\frac{1}{8} \) after appropriate substitution, identifying the function's behavior at \( x=2 \).

Factorization in Limits

Factorization simplifies complex expressions and is a crucial technique in solving limits. By breaking down polynomials into simpler components, we can cancel common terms from numerators and denominators when applicable. This helps resolve indeterminate forms in limits.

• Identify and factor both the numerator and denominator if possible. Look for common factors that can cancel out.
• Once canceled, analyze the remaining expression to evaluate the limit accurately.

In this exercise, we transformed \( \frac{x^3 - 5x^2 + 6x}{x^4 - 4x^2} \) into \( \frac{x(x-3)}{x^2(x+2)} \) after cancellation. Factorization was instrumental in simplifying the expression and clarifying the limit behavior as \( x \) approached specific values. This step is crucial in simplifying and accurately evaluating limits.