Problem 28 - Calculus Early Transcendentals

Question

Prove Theorem 2.2: If \(f(x)=m x+b,\) then \(\lim _{x \rightarrow a} f(x)=m a+b\) for constants \(m \neq 0\) and \(b\) (the case \(m=0\) is considered in Exercise \(27 \mathrm{a}\) ). (Hint: For a given \(\varepsilon>0, \text { let } \delta=\varepsilon /|m| .)\) Explain why this result implies that linear functions are continuous.

Step-by-Step Solution

Answer

Question: Prove that for any linear function, \(f(x) = mx + b\), the limit as x approaches a, is equal to \(ma + b\), and explain why this implies that linear functions are continuous. Answer: Following the epsilon-delta definition of a limit, we showed that for our choice of \(\delta = \varepsilon / |m|\), whenever \(0 < |x-a| < \delta\), we have \(|f(x) - (ma+b)| < \varepsilon\). This verifies that the limit as x approaches a is equal to \(ma + b\). Since the limit as x approaches a is equal to the function's value at that point, linear functions are continuous.

Step 1: Understanding the precise definition of a limit

The precise definition of a limit states: $$\lim _{x \rightarrow a} f(x)=L$$ if for every \(\varepsilon > 0\), there exists a \(\delta>0\) such that whenever \(0<|x-a|<\delta,\) then \(|f(x)-L|<\varepsilon\). Our goal is to find a \(\delta\) that satisfies the above condition for our function \(f(x) = mx + b\).

Step 2: Apply the definition to the function

Given \(\varepsilon > 0\), we want to find a \(\delta\) such that if \(0<|x-a|<\delta\), then \(|f(x) - (ma+b)| < \varepsilon\). Our goal is to manipulate \(f(x) - (ma + b)\) to determine a proper \(\delta\). Let's compute the difference \(|f(x) - (ma+b)|\): \(|f(x) - (ma+b)| = |(mx+b) - (ma+b)| = |m(x-a)| = |m||x-a|\) Now, we want to find a suitable \(\delta\) such that \(|m||x-a| < \varepsilon\).

Step 3: Determine the delta

From the hint given in the exercise, let's choose \(\delta = \varepsilon / |m|\). So, if \(0 < |x-a| < \delta\), then \(|m||x-a| < |m|(\varepsilon/|m|) = \varepsilon\)

Step 4: Showing that delta satisfies the condition

Now, we have shown that, for our choice of \(\delta = \varepsilon / |m|\), whenever \(0 < |x-a| < \delta\), we have \(|f(x) - (ma+b)| < \varepsilon\). This completes the proof, as it verifies the limit definition. Thus, $$\lim _{x \rightarrow a} f(x)=ma + b$$

Step 5: Linear functions are continuous

A function is continuous at a point \(a\) if and only if \(\lim_{x \rightarrow a} f(x) = f(a)\). Since we showed that for a linear function, the limit as \(x\) approaches \(a\) is equal to \(ma+b\), which is just \(f(a)\), we can conclude that linear functions are continuous.

Key Concepts

Continuity of Functions

Continuity in mathematics refers to how smoothly a function behaves as its input varies. For a function to be considered continuous at a point, the function must not have any sudden jumps or breaks. In simpler terms, imagine drawing the graph of a function with a pen. If you can draw it without lifting your pen off the paper, the function is continuous.
In precise mathematical language, a function is continuous at a point \(a\) if the limit of the function as \(x\) approaches \(a\) is equal to the function's value at \(a\). This can be expressed as:

• \(\lim_{x \rightarrow a} f(x) = f(a)\)

For linear functions, such as \(f(x) = mx + b\), this condition holds truly because these functions are a straight line without any disturbances. As a result, the limit at any point is equal to the function's value at that point, solidifying the concept that linear functions are inherently continuous.

Epsilon-Delta Definition of Limit

The epsilon-delta definition is foundational in understanding limits and describes exactly what it means for a function to have a limit at a point. Let's break it down step by step. This definition asserts:

• For every \(\varepsilon > 0\) (epsilon representing any arbitrarily small number), there exists a \(\delta > 0\) (delta being another small number),
• such that whenever \(0 < |x-a| < \delta\), the absolute difference between the function's value and the limit is less than epsilon: \(|f(x) - L| < \varepsilon\).

This approach provides a rigorous method to prove limits. In the exercise, you applied this to the linear function \(f(x)=mx+b\) by determining a proper \(\delta\) (suggested as \(\delta = \varepsilon / |m|\)). This choice of delta ensures that as \(x\) gets very close to \(a\) (within \(\delta\)), \(f(x)\) gets very close to \(ma+b\) (within \(\varepsilon\)). This shows the power of the epsilon-delta definition in proving the behavior of functions around a point.

Linear Functions

Linear functions are among the simplest yet most important forms of mathematical functions. A linear function is expressed as \(f(x) = mx + b\), where \(m\) and \(b\) are constants. The number \(m\) represents the slope, which tells us how steep the line is, while \(b\) corresponds to the y-intercept, indicating where the line crosses the y-axis.
Linear functions depict a straight-line graph, which means they have constant rates of change. This uniform behavior makes them predictable and widely used in various fields, like physics, economics, and everyday decision-making, to understand and model constant trends.
In terms of limits and continuity, linear functions shine due to their straightforward nature. Because they are continuous, finding limits at any point along their line is effortless. The continuous nature of linear functions means you can calculate the limit as \(x\) approaches a point simply by plugging \(x\) into the function, maintaining consistency with the function's value at that point. This inherent simplicity is what makes linear functions a perfect example when first learning about limits and continuity in calculus.