Problem 4 - Single Variable Calculus

Question

You are given \(\lim _{x \rightarrow a} f(x)=L\) and a tolerance \(\varepsilon\). Find a number \(\delta\) such that \(|f(x)-L|<\varepsilon\) whenever \(0<|x-a|<\delta\) \(\lim _{x \rightarrow-2}(3 x-2)=-8 ; \quad \varepsilon=0.05\)

Step-by-Step Solution

Answer

The short answer is: \(\delta \approx 0.0167\).

Step 1: Identify the function's limit

We are given that \(\lim_{x \rightarrow -2} (3x - 2) = -8\). This means that as \(x\) approaches \(-2\), the value of the function \((3x - 2)\) approaches \(-8\).

Step 2: Identify the Tolerance

We are given a tolerance \(\varepsilon = 0.05\). This means we want to find a \(\delta\) such that the difference between the function's value and its limit is less than \(0.05\).

Step 3: Write down the condition for \(\delta\)

We need to find a \(\delta\) such that \(|f(x) - L| < \varepsilon\) whenever \(0 < |x - a| < \delta\). In this case, we have: - \(f(x) = 3x - 2\) - \(L = -8\) - \(a = -2\) - \(\varepsilon = 0.05\) Thus, we want to find a \(\delta\) such that \(|(3x - 2) - (-8)| < 0.05\) whenever \(0 < |x - (-2)| < \delta\).

Step 4: Simplify the Condition

To simplify the condition, let's first simplify the absolute value expression: \(|(3x - 2) - (-8)| = |3x - 2 + 8| = |3x + 6|\) Now the condition becomes: \(|3x + 6| < 0.05\) whenever \(0 < |x + 2| < \delta\)

Step 5: Find the appropriate \(\delta\)

Now we want to find a \(\delta\) value such that \(|3x + 6| < 0.05\) whenever \(0 < |x + 2| < \delta\). We can reason as follows: - Since \(|x + 2| < \delta\), we can multiply both sides of the inequality by \(3\) to get \(|3x + 6| < 3\delta\). - We want \(|3x + 6| < 0.05\), so we can set \(3\delta = 0.05\). - Solving for \(\delta\), we have \(\delta = \frac{0.05}{3} \approx 0.0167\). Thus, our desired \(\delta\) value is \(\delta \approx 0.0167\).

Key Concepts

Epsilon-Delta Definition

The concept of the **epsilon-delta definition** is crucial when understanding limits in calculus. It's all about defining how close we need to be to a particular point. In essence, when we say \( \lim_{x \rightarrow a} f(x) = L \), we mean:

• For every positive number \( \varepsilon \), no matter how small,
• There exists a positive number \( \delta \) such that
• Whenever \( |x - a| < \delta \), it ensures \( |f(x) - L| < \varepsilon \)

The \( \delta \) defines how close \( x \) must be to \( a \), ensuring \( f(x) \) is as close to \( L \) as needed. This definition solidifies our understanding of reaching the limit without exactly touching it.

Tolerance in Limits

Understanding **tolerance in limits** is about recognizing how much deviation from the limit is acceptable. The value \( \varepsilon \) represents this tolerance.

• It dictates just how close \( f(x) \) needs to get to \( L \).
• A smaller \( \varepsilon \) demands more precision, meaning \( f(x) \) must be even closer to \( L \).

When working problems, we often choose an \( \varepsilon \) to determine the necessary accuracy in calculations. In practical terms, \( \varepsilon \) determines the 'wiggle room' we have around the limit \( L \) which is especially crucial in engineering and applied sciences.

Finding Delta in Limits

**Finding delta in limits** involves determining how closely \( x \) must approach \( a \) to keep \( f(x) \) within the designated \( \varepsilon \) range of \( L \). This critical step ensures our function remains within that "tolerated" difference.

• We identify \( |x - a| < \delta \) which maintains \( |f(x) - L| < \varepsilon \).
• Essentially, \( \delta \) controls the input closeness needed for output precision.

To find \( \delta \), sometimes, we need algebraic manipulations as seen in the exercise: solving \( |3x + 6| < 3\delta \) leads to \( \delta = \frac{0.05}{3} \). Analytical skills in this step bridge the input-output proximity.

Absolute Value Inequality

Using **absolute value inequality** is a method to express the range within which the function must stay close to the limit. It provides a flexible way of showing proximity both above and below the limit.

• The expression \( |f(x) - L| < \varepsilon \) implies strict bounds around \( L \).
• It's a two-sided inequality handling positive and negative deviations.

To solve, break down the absolute value into straightforward expressions. For instance, solving \(|3x + 6| < 0.05\) gives the necessary control over how \( f(x) \) measures up to \( L \) within \( \varepsilon \). This technique is a backbone for precision in calculus problems.

Continuity of Functions

The **continuity of functions** reflects the idea that a function should not have abrupt shifts or breaks around a point \( a \). If we have \( \lim_{x \rightarrow a} f(x) = f(a) \), it indicates continuity at point \( a \).

• Continuous functions allow limits to match function values neatly.
• They do not break or jump unexpectedly at \( x = a \).

In the epsilon-delta context, continuous functions guarantee that our \( f(x) \) stays nice and smooth as \( x \) zooms in on \( a \). Understanding continuity is essential because it ensures the function behaves predictably near the point of interest, making limit calculations simpler and more reliable.