Problem 39 - Calculus of a Single Variable

Question

Using the \(\varepsilon-\delta\) Definition of Limit In Exercises \(37-48\) , find the limit \(L\) . Then use the \(\varepsilon-\delta\) definition to prove that the limit is \(L .\) $$ \lim _{x \rightarrow-4}\left(\frac{1}{2} x-1\right) $$

Step-by-Step Solution

Answer

The limit \(L\) was found to be -3 and this was confirmed using the \(\varepsilon-\delta\) definition of a limit.

Step 1: Calculate Limit L

As the function is simply \(f(x) = \frac{1}{2} x -1\), a continuous function, we can find the limit by directly substituting the value \(x = -4\) into the function, this will give us \(L = \frac{1}{2}*(-4) -1 = -3\). So, our limit \(L\) is \(-3\).

Step 2: Understand the epsilon-delta definition

The \(\varepsilon-\delta\) definition states that a limit \(L\) of a function at a certain point exists if for every number \(\varepsilon > 0\), there is a number \(\delta > 0\) such that if the absolute difference between \(x\) and the point is less than \(\delta\) (except at the point itself), then the absolute difference between \(f(x)\) and \(L\) will be less than \(\varepsilon\). We have to use this definition to prove our limit.

Step 3: Apply the epsilon-delta proof

Our function is \(f(x) = \frac{1}{2} x -1\) and we're looking at the limit as \(x\) approaches \(-4\). We want to show that for any given \(\varepsilon > 0\), we can find a \(\delta > 0\) such that if \(0 < |x+4| < \delta\), then \( |\frac{1}{2}x -1 -(-3)| < \varepsilon\). Simplifying the inequality inside the absolute value gives us \(|\frac{1}{2}x + 2| < \varepsilon\). Dividing the absolute expression and \(\varepsilon\) by \(\frac{1}{2}\), we get \(|x+4| < 2 \varepsilon\). Thus, we can choose \(\delta\) to be \(2 \varepsilon\) to satisfy the \(\varepsilon-\delta\) definition of the limit. Therefore, we have proven that the limit of the function \(f(x)\) as \(x\) approaches \(-4\) is indeed \(-3\) by the \(\varepsilon-\delta\) proof.

Key Concepts

Limit of a Function

The concept of a limit is foundational in calculus. It uses the idea of approaching something without necessarily reaching it. To find the limit of a function as the variable approaches a certain value, we look at the values that the function approaches as the input gets closer and closer to that point. For instance, if we want the limit of \( f(x) \) as \( x \) approaches some number, we're interested in what \( f(x) \) becomes as \( x \) gets increasingly close to that number from either side.
For the function \( f(x) = \frac{1}{2}x - 1 \), to find \( \lim_{x \to -4} f(x) \), substitute \( x = -4 \) into the function. This makes it much simpler due to the linear nature of our function. Acting as a real number function, it means \( f(x) \) tends not to "do anything weird" around \( x = -4 \). Instead, it smoothly approaches a value. From our example, by substituting, we get \(-3\) as the limit.

Continuity

Continuity in a function means there are no gaps, jumps, or unmanageable spikes at any point within its domain. Formally, a function \( f(x) \) is continuous at a point \( c \) if the limit of \( f(x) \) as \( x \) approaches \( c \) is exactly \( f(c) \).
This property is crucial because, for continuous functions, direct substitution is often a valid method to find limits. Linear functions, like \( f(x) = \frac{1}{2}x - 1 \), are continuous across the real numbers. At \( x = -4 \), the function is continuous, so \( \lim_{x \to -4} f(x) \) is just \( f(-4) = -3 \). Continuous behavior ensures that as you trace along the graph of the function, you pass through every vertical section seamlessly.

Proof of Limit

The \( \varepsilon-\delta \) definition is a formal way to establish limits. It works by tackling both the function's behavior near a specific point and ensuring that we can get as close as needed to the limit value. According to this definition, a limit \( L \) exists if, for every tiny distance \( \varepsilon > 0 \), there's a smaller range \( \delta > 0 \), which confines \( x \) close enough to make \( |f(x) - L| < \varepsilon \).
Applying this to the function \( f(x) = \frac{1}{2}x - 1 \) and \( x \to -4 \), we need \(|\frac{1}{2}x + 2| < \varepsilon\). Solving for \( \delta \), choosing it as \( 2\varepsilon \) suffices, forming a bridge between \( x \) and the limit \( L = -3 \). This process proves rigorously that the function does approach \( -3 \) as described by the definition.

Calculus

Calculus is the mathematical study of continuous change. Among its main ideas are concepts like limits, differentiation, and integration. When beginning with limits, such as our limit definition discussed, we take the first step into this vast domain.
Calculus helps in exploring functions beyond their explicit outputs by predicting behavior around points and overall trends. Whether determining speed from a distance function or understanding local behavior in various phenomena, limits and continuity form the bedrock. The epsilon-delta definition is a testament to the rigorous tools calculus provides to ensure that we understand function behaviors precisely, setting the stage for further learning on derivatives and integrals.