Problem 57 - Thomas Calculus

Question

Find the limits in Exercises \(53-58\). $$\lim \frac{x^{2}-3 x+2}{x^{3}-2 x^{2}}\mathrm{as}$$ \begin{equation} \begin{array}{ll}{\text { a. }} & {x \rightarrow 0^{+}} \quad\quad\quad\quad\quad\quad\quad {\text { b. } x \rightarrow 2^{+}} \\\ {\text { c. }} & {x \rightarrow 2^{-}} \quad\quad\quad\quad\quad\quad\quad {\text { d. } x \rightarrow 2} \\ {\text { e. }} & {\text { What, if anything, can be said about the limit as } x \rightarrow 0 ?}\end{array} \end{equation}

Step-by-Step Solution

Answer

a. \(+\infty\), b. \(\frac{1}{4}\), c. \(\frac{1}{4}\), d. \(\frac{1}{4}\), e. Limit doesn't exist.

Step 1: Analyze the Limit as x Approaches 0 from the Right

Initially substitute \(x = 0^{+}\) into \(\frac{x^2 - 3x + 2}{x^3 - 2x^2}\). The numerator becomes \(0^2 - 3 \cdot 0 + 2 = 2\) and the denominator becomes \(0^3 - 2 \cdot 0^2 = 0\). Since the denominator is 0, observe the sign of the limit: for \(x \to 0^{+}\), \(x^2\) is positive and \(3x\) goes to 0. Thus, the limit as \(x \rightarrow 0^+\) is positive infinity, or \(+\infty\).

Step 2: Analyze the Limit as x Approaches 2 from the Positive Side

Substitute \(x = 2^{+}\) into the expression: the numerator becomes \((2^2) - (3 \cdot 2) + 2 = 4 - 6 + 2 = 0\), and the denominator \(2^{3} - 2(2^2) = 8 - 8 = 0\). Since both the numerator and denominator approach zero, apply L'Hôpital's Rule. Differentiate the numerator \(2x - 3\) and the denominator \(3x^2 - 4x\). Evaluate \(\lim_{x \to 2^+} \frac{2x-3}{3x^2-4x} = \frac{2(2) - 3}{3(2)^2 - 4(2)} = \frac{1}{4} \).

Step 3: Analyze the Limit as x Approaches 2 from the Negative Side

As \(x \rightarrow 2^-\), both the numerator and denominator also approach zero. By continuity in derivatives, use L'Hôpital's Rule again. As calculated in Step 2, \(\lim_{x \to 2^-} \frac{2x-3}{3x^2-4x} = \frac{1}{4}\).

Step 4: Analyze the Limit as x Approaches 2

Since both \(\lim_{x \to 2^+}\) and \(\lim_{x \to 2^-}\) are \(\frac{1}{4}\), the two one-sided limits at \(x = 2\) agree. Therefore, \(\lim_{x \to 2} \frac{x^2 - 3x + 2}{x^3 - 2x^2} = \frac{1}{4} \).

Step 5: Conclusion on the Limit as x Approaches 0

Since the limit as \(x \to 0^{+}\) is \(+\infty\) and the expression is undefined at \(x = 0\), the limit as \(x \rightarrow 0\) does not exist overall, but \(x \rightarrow 0^{+}\) is \(+\infty\).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool for evaluating limits that result in indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). This rule states that if the limits of both the numerator and the denominator of a fraction approach zero or infinity, the limit of the quotient of their derivatives will be equal to the limit of the original fraction. The rule can be expressed as: \[\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}\]provided both \(f'(x)\) and \(g'(x)\) are non-zero near \(c\). This helps to simplify otherwise complicated indeterminate forms. For instance, in the given problem, when evaluating the limit as \(x\) approaches 2 from both sides, we first obtain \(\frac{0}{0}\). By differentiating the numerator and the denominator, we simplify the expression to \(\frac{2x - 3}{3x^2 - 4x}\) and find the values using these derivatives, resulting in a precise solution.

One-Sided Limits

One-sided limits are used to determine the behavior of a function as it approaches a specific point from either the left or the right. These are written as \(\lim_{x \to c^-} f(x)\) for the left-hand limit and \(\lim_{x \to c^+} f(x)\) for the right-hand limit. They are particularly useful when dealing with functions that are not continuous at a point or exhibit different behaviors on either side of a point.

• Left-Hand Limit: What the function is approaching as it comes from values less than the point of interest.
• Right-Hand Limit: What the function is approaching as it comes from values greater than the point of interest.

In the exercise provided, one-sided limits were calculated for \(x \to 2^+\) and \(x \to 2^-\), where both resulted in \(\frac{1}{4}\), indicating that the function is smooth and continuous at \(x = 2\).

Continuous Functions

Continuous functions are those that do not have any abrupt changes in value, meaning they can be drawn without lifting the pen off the paper. A function \(f(x)\) is continuous at a point \(c\) if the following three conditions are met:

• \(f(c)\) is defined.
• \(\lim_{x \to c} f(x)\) exists.
• \(\lim_{x \to c} f(x) = f(c)\).

In the step-by-step solution, continuity plays a crucial role in determining the limit as \(x\) approaches 2. Because both the left and right-hand limits at \(x = 2\) are equal, the limit exists and the function is continuous at this point. This helps confirm that there are no sudden jumps or breaks at \(x = 2\), leading to the conclusion that the limit is consistently \(\frac{1}{4}\).

Infinity Limits

Infinity limits describe the behavior of a function as the input approaches very large values, either positively or negatively. These are written as \(\lim_{x \to \infty} f(x)\) or \(\lim_{x \to -\infty} f(x)\). It often signifies a function's asymptotic behavior, which might not land on a specific number.In the exercise, the focus was on one-sided infinite limits as \(x \to 0^+\). When approaching 0 from the positive side, the original denominator approached 0 from the positive side, leading to a vertical asymptote and a resulting limit of \(+\infty\). This is a classic example of an infinite limit where the function heads towards positive infinity rather than a fixed numerical value as it draws near the point from one side. Recognizing these patterns is vital as it signifies the potential of a function growing indefinitely.