Problem 1 - Single Variable Calculus

Question

In Exercises 1-10 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 \(0<|x-a|<\delta\) \(\lim _{x \rightarrow 2} 3 x=6 ; \quad \varepsilon=0.01\)

Step-by-Step Solution

Answer

We are given \(\lim_{x \rightarrow 2} 3x = 6\) and need to find a \(\delta\) such that \(|f(x)-6|<0.01\) whenever \(0<|x-2|<\delta\). By substituting the function \(f(x) = 3x\) and manipulating the inequality, we find that \(\delta = \frac{0.01}{3}\).

Step 1: Understand the problem statement

We are given that \(\lim_{x \rightarrow 2} 3x = 6\) and we need to find a \(\delta\) such that \(|f(x)-6|<0.01\) whenever \(0<|x-2|<\delta\).

Step 2: Substitute the function and simplify

In our case, \(f(x) = 3x\), so \[|f(x)-6| = |3x-6|\]

Step 3: Manipulate the inequality

We want to show that \(|3x-6|<0.01\) whenever \(0<|x-2|<\delta\). First, let's divide both sides of the inequality by 3: \[\frac{|3x-6|}{3} = |x-2| < \frac{0.01}{3}\]

Step 4: Find delta

From step 3, we know that \(|x-2| < \frac{0.01}{3}\) whenever \(|3x-6|<0.01\). So, we can take \(\delta = \frac{0.01}{3}\). Thus, a number \(\delta\) such that \(|f(x)-6|<0.01\) whenever \(0<|x-2|<\delta\) is \(\delta = \frac{0.01}{3}\).

Key Concepts

Epsilon-delta definition

The epsilon-delta definition is a formal way of defining the limit of a function. This definition asserts that a function, say \( f(x) \), approaches a limit \( L \) as \( x \) approaches \( a \), if for every tiny positive number \( \varepsilon \), there exists another tiny positive number \( \delta \), such that whenever \( x \) is within \( \delta \) units of \( a \) (but not equal to \( a \)), \( f(x) \) stays within \( \varepsilon \) units of \( L \).
This can be written mathematically as:

• For every \( \varepsilon > 0 \), there exists a \( \delta > 0 \)
• Such that if \( 0 < |x - a| < \delta \), then \( |f(x) - L| < \varepsilon \)

Using this definition allows us to rigorously prove whether a function truly approaches a certain value as the input approaches a selected point in the domain. Here, understanding the epsilon-delta concept is crucial for accurately determining the behavior of functions around specific points.

Continuity

Continuity at a point ensures a smooth and unbroken behavior of a function at that point. For a function \( f(x) \) to be continuous at a point \( a \):

• \( f(a) \) must be defined.
• \( \lim_{x \to a} f(x) \) exists and equals \( f(a) \).

If these conditions are met, the function does not have any abrupt jumps, holes, or breaks at that specific point. This means tracing the graph of the function around \( a \) should feel seamless and uninterrupted.
Continuity is essential for many mathematical concepts because it confirms that small changes in the input around \( a \) lead to small and predictable changes in the output. This ensures stability and predictability in calculations and applied scenarios, making understanding continuity crucial for studying functions and their limits.

Inequalities in calculus

Inequalities are fundamental in calculus, especially when working with limits and approximations. They allow us to form bounds, showing how close a function's value can get to a specific limit or point.
In the context of limits, inequalities like \( |f(x) - L| < \varepsilon \) establish a permissible range (\( \varepsilon \)) for how much \( f(x) \) can deviate from \( L \). Using inequalities, we can manipulate and simplify expressions to find tolerances, such as \( \delta \), that satisfy the given conditions.
To effectively solve calculus problems, understanding and applying inequalities are key. They make it possible to appropriately scale and adjust values, ensuring the function behaves within expected boundaries. In the given exercise, inequalities help us bridge the gap between a function’s algebraic form and its behavior as inputs approach a crucial point.