Problem 17 - Calculus: Early Transcendentals

Question

Finding a symmetric interval Let \(f(x)=\frac{2 x^{2}-2}{x-1}\) and note that \(\lim _{x \rightarrow 1} f(x)=4 .\) For each value of \(\varepsilon,\) use a graphing utility to find all values of \(\delta>0\) such that \(|f(x)-4|<\varepsilon\) whenever \(0<|x-1|<\delta\) a. \(\varepsilon=2\) b. \(\varepsilon=1\) c. For any \(\varepsilon>0,\) make a conjecture about the value of \(\delta\) that satisfies the preceding inequality.

Step-by-Step Solution

Answer

Question: For a given function \(f(x) = \frac{2x^2 - 2}{x - 1}\), find the values of \(\delta\) for which \(|f(x) - 4| < \varepsilon\) holds true when \(0 < |x - 1| < \delta\) for three cases: \(\varepsilon = 2\), \(\varepsilon = 1\), and make a conjecture for any \(\varepsilon > 0\). Answer: Analyzing the graph of the function, the possible values of \(\delta\) can be determined by finding the minimum distance between the intersection points of the function and the horizontal lines \(y = 4 \pm \varepsilon\) to \(x = 1\). As \(\varepsilon\) becomes smaller, the corresponding range of \(x\)-values and the allowed \(\delta\) values become narrower and symmetric around \(x = 1\).

Step 1: Understanding the function around x = 1

Using the algebraic limit, we find that \(\lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} \frac{2x^2 - 2}{x - 1} = 4\). In order to find appropriate \(\delta\) values for given \(\varepsilon\) values, we will analyze the function graphically and find the regions where the inequality \(|f(x) - 4| < \varepsilon\) is true.

Step 2: Finding delta values for epsilon = 2

Let's use a graphing utility to analyze the region around \(x = 1\) where \(|f(x) - 4| < 2\). Visualizing the graph for the given function and the horizontal bounds \(y = 6\) and \(y = 2\), we can find the interval of \(x\) values satisfying the inequality. Mark the points of intersection and find their \(x\)-coordinates, except for \(x = 1\). The range of these x-values will be used to determine \(\delta\). For this case, there would be multiple intervals in which the inequality holds true, so there would be multiple possible values of \(\delta>0\).

Step 3: Finding delta values for epsilon = 1

This time, we will consider the region around \(x = 1\) where \(|f(x) - 4| < 1\). Visualizing the graph for the given function and the horizontal bounds \(y = 5\) and \(y = 3\), we can find a narrower interval of \(x\) values satisfying the inequality compared to \(\varepsilon = 2\). Follow the same procedure to determine \(\delta\) values for this case.

Step 4: Conjecture for any epsilon > 0

Based on the values found for \(\delta\) in the cases where \(\varepsilon = 2\) and \(\varepsilon = 1\), we can notice that as \(\varepsilon\) becomes smaller, the corresponding range of \(x\)-values (and thus the allowed \(\delta\) values) becomes narrower, symmetrically around \(x = 1\). As a conjecture, for any \(\varepsilon > 0\), we can estimate the \(\delta\) value by analyzing the graph of the function and observing the intersection points with the horizontal lines at \(y = 4 \pm \varepsilon\). Find the \(x\)-coordinates of these points, and the minimum distance from these points to \(x = 1\) will give a symmetric interval around \(x = 1\) for the chosen \(\varepsilon\).

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition is a fundamental concept in calculus used to rigorously define the limit of a function. It provides a precise way to describe what it means for a function to approach a certain value as the input approaches a specified point.

The definition revolves around two small positive numbers, \(\varepsilon\) (epsilon) and \(\delta\) (delta). We say that \(\lim_{{x \to a}} f(x) = L\) if, for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that, whenever \(0 < |x - a| < \delta\), it follows that \(|f(x) - L| < \varepsilon\).

• The \(\varepsilon\) represents how close \(f(x)\) should be to \(L\).
• The \(\delta\) indicates how close \(x\) must be to \(a\) to meet the \(\varepsilon\) condition.
• The goal is to find a suitable \(\delta\) for each \(\varepsilon\).

This definition ensures that the function gets as close as desired to the limit value, ensuring precision and stability.

Graphical Analysis of Functions

Graphical analysis in calculus helps visualizing and understanding the behavior of functions around specific points, like detecting limits visually.

With a given function like \( f(x) = \frac{2x^2 - 2}{x-1} \) and a point of interest, say \( x = 1 \), graphing the function can help identify values where the function is near a specific limit like 4.

• The graph provides a visual cue to where \(|f(x) - 4| < \varepsilon\). This helps determine the intervals of \(x\) satisfying this condition.
• By marking the intersections between the function and horizontal lines at \(y = 4 \pm \varepsilon\), one can find the \(x\)-values that work for the epsilon condition.

Graphing utilities like computer software or graphing calculators are typically used for this analysis, enabling you to zoom in on the areas of interest. Observing how function curves approach horizontal lines assists in deriving precise values for \(\delta\).

Symmetric Intervals

Symmetric intervals are used to ensure that the \(\delta\) region around a point remains even and balanced, usually around the point where the limit is evaluated.

When working to find \(\delta\) values as in our case with \(f(x)\), determining \(\delta\) involves finding an interval \(|x - 1| < \delta\). This implies symmetry around \(x = 1\).

• The intervals indicate that the acceptable \(x\) values are the same distance away both left and right from \(x = 1\).
• This approach ensures that \(|f(x) - 4| < \varepsilon\) within the symmetric bounds.
• Smaller values of \(\varepsilon\) create smaller, tighter intervals.

Symmetric intervals thus help maintain a consistent condition for limits throughout the calculus problem, providing a strong visual and analytical cue for evaluating functions precisely.