Problem 42 - 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 \(L .\) $$ \lim _{x \rightarrow 2}(-1) $$

Step-by-Step Solution

Answer

The limit is -1. For the \(\varepsilon - \delta\) definition, for every \(\varepsilon > 0\), choose \(\delta = 1\). This secures that \(|-1 - (-1)| = 0 < \varepsilon\) and hence proves the limit of the constant function is the constant itself, regardless of the \(x\) value it approaches.

Step 1: Find the Limit \(L\)

Since the function here is a constant (-1), regardless of what value \(x\) approaches, the limit is simply the constant itself. So, the limit \(L = -1\).

Step 2: Apply the \(\varepsilon-\delta\) Definition

The \(\varepsilon-\delta\) definition of limit states that for every real number \(\varepsilon > 0\), there must exist a real number \(\delta > 0\) such that whenever \(0 < |x - 2| < \delta\), the inequality \(|-1 - L| < \varepsilon\) must hold. In this case, given \(L = -1\), the inequality would always be \(| -1 - (-1) | = 0 < \varepsilon\), regardless of the value of \(x\). As such, for any \(\varepsilon > 0\), choose \(\delta = 1\). Then, whenever \(0 < |x - 2| < 1\), we will have \(|-1 - (-1)| = 0 < \varepsilon\). This shows that the \(\varepsilon-\delta\) definition holds and thus proves the limit.

Key Concepts

Limits in Calculus

In calculus, limits are essential for understanding the behavior of functions as they approach specific points or even infinity. A limit describes the value that a function ‘approaches’ as the input (or independent variable) approaches some value. For instance, the expression \( \lim_{x \to a} f(x) \) means that as \( x \) gets arbitrarily close to \( a \) from both sides of the real number line, the value of the function \( f(x) \) gets arbitrarily close to a certain number, which is the limit.

The concept of limits allows us to handle situations where the function might not actually be defined at the point \( a \) itself, yet it has a predictable trend as we draw near to that point. Limits form the foundation for other important calculus topics, including derivatives and integrals, which describe change and accumulation, respectively.

Proof of Limits

Proving limits rigorously often involves the \( \varepsilon-\delta \) definition, which sets the stage for a formal, mathematical understanding of limits. The \( \varepsilon-\delta \) criterion for the limit says that for a function \( f(x) \) with a limit \( L \) as \( x \) approaches \( a \) can be proven if for every positive number \( \varepsilon > 0 \) there exists a corresponding positive number \( \delta > 0 \) such that if \( 0 < |x - a| < \delta \) then \( |f(x) - L| < \varepsilon \).

In simpler terms, this means no matter how small a distance \( \varepsilon \) we set around the limit \( L \) on the y-axis, we can find a distance \( \delta \) on the x-axis so that when \( x \) is within this \( \delta \) distance from \( a \) (but not equal to \( a \)), \( f(x) \) is within the \( \varepsilon \) distance from \( L \). Using this definition allows mathematicians to confirm the behavior of functions near certain points in a way that is precise and can be universally agreed upon.

Constant Function Limit

The limit of a constant function is a unique case because it's delightfully straightforward. For any constant function \( f(x) = c \) where \( c \) is a constant, the limit as \( x \) approaches any value \( a \) is always \( c \). The \( \varepsilon-\delta \) proof of this limit is also quite simple, because the value of the function does not change no matter what \( x \) is; thus, it’s always exactly \( c \) units away from the limit \( L \), which is itself \( c \).

For example, if we have \( \lim_{x \to 2} (-1) \) as in our exercise, we know right away that the limit is \( -1 \) since the function we're looking at is simply \( f(x) = -1 \). The \( \varepsilon-\delta \) proof comes out neatly because \( |f(x) - L| = |-1 - (-1)| = 0 < \varepsilon \) for any \( \varepsilon > 0 \) we choose; thus, \( L = -1 \) is indeed the limit. In effect, proving the limit of a constant function comes down to validating the understanding that a constant doesn't change—it's always itself, hence the term 'constant.'