Problem 21
Question
Use the formal definition of limits to prove each statement. \(\lim _{x \rightarrow c}(m x)=m c\), where \(m\) is a constant
Step-by-Step Solution
Verified Answer
The limit \( \lim_{x \rightarrow c}(mx)=mc \) is proven by the definition of limits using \( \delta = \frac{\varepsilon}{|m|} \).
1Step 1: Recall the Formal Definition of a Limit
According to the formal definition, \( \lim_{x \to c} f(x) = L \) means that for every \( \varepsilon > 0 \) there exists a \( \delta > 0 \) such that if \( 0 < |x - c| < \delta \), then \( |f(x) - L| < \varepsilon \). Our goal is to apply this definition to prove that \( \lim_{x \to c}(mx) = mc \).
2Step 2: Write the Target Inequality
The target inequality based on \( f(x) = mx \) and \( L = mc \) is \( |mx - mc| < \varepsilon \). The proof will involve showing that we can find a \( \delta \) that satisfies this condition whenever \( 0 < |x - c| < \delta \).
3Step 3: Simplify the Absolute Difference
Notice that \( |mx - mc| = |m(x - c)| \). Because the absolute value of a product is the product of the absolute values, we have \( |m||x - c| < \varepsilon \).
4Step 4: Choose \( \delta \) Appropriately
To ensure \( |m||x - c| < \varepsilon \), set \( \delta = \frac{\varepsilon}{|m|} \) (assuming \( m eq 0 \)). Therefore, if \( 0 < |x - c| < \delta \), then \( |mx - mc| = |m||x - c| < |m|\delta = \varepsilon \).
5Step 5: Conclude the Proof
By choosing \( \delta = \frac{\varepsilon}{|m|} \), we have shown that for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that \( 0 < |x - c| < \delta \) implies \( |mx - mc| < \varepsilon \). Therefore, the limit \( \lim_{x \to c} (mx) = mc \) is proven.
Key Concepts
Formal Definition of LimitsProof of LimitsEpsilon-Delta Definition
Formal Definition of Limits
When we talk about limits in calculus, we're often referring to how a function behaves as it approaches a certain point, even if it never quite reaches that point. The formal definition of a limit adds precision to this idea. According to this definition, we say that \( \lim_{x \to c} f(x) = L \) if, for every positive number \( \varepsilon \), no matter how small, there is a corresponding positive number \( \delta \) such that whenever \( 0 < |x - c| < \delta \), it follows that \( |f(x) - L| < \varepsilon \).
This definition helps us rigorously show how close \( f(x) \) can get to an expected value \( L \) as \( x \) gets closer to \( c \). The challenge lies in finding the right \( \delta \) to match any \( \varepsilon \) we might choose. This interplay between \( \varepsilon \) and \( \delta \) ensures that our function \( f(x) \) is approaching \( L \) closely, as \( x \) approaches \( c \). This precision is crucial for many proofs and applications in calculus.
This definition helps us rigorously show how close \( f(x) \) can get to an expected value \( L \) as \( x \) gets closer to \( c \). The challenge lies in finding the right \( \delta \) to match any \( \varepsilon \) we might choose. This interplay between \( \varepsilon \) and \( \delta \) ensures that our function \( f(x) \) is approaching \( L \) closely, as \( x \) approaches \( c \). This precision is crucial for many proofs and applications in calculus.
- \( \varepsilon \) represents the "closeness" we desire for \( f(x) \) to \( L \).
- \( \delta \) represents the "wiggle room" around \( c \) such that \( f(x) \) stays within the \( \varepsilon \) boundary of \( L \).
Proof of Limits
Proving limits using the formal definition can seem challenging at first, but it actually follows a logical step-by-step process. Let's take the limit \( \lim_{x \to c}(mx) = mc \) as an example where \( m \) is a constant.
The goal of the proof is to show that for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that when \( 0 < |x - c| < \delta \), the inequality \( |mx - mc| < \varepsilon \) holds.
Here's how the proof unfolds:
The goal of the proof is to show that for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that when \( 0 < |x - c| < \delta \), the inequality \( |mx - mc| < \varepsilon \) holds.
Here's how the proof unfolds:
- First, express the target inequality: \( |mx - mc| < \varepsilon \).
- Simplify this to \( |m(x - c)| < \varepsilon \).
- Recognize \( |m| |x - c| < \varepsilon \). Here, we're essentially scaling \( |x - c| \) by \( |m| \), the absolute value of our constant.
- To find \( \delta \), set \( \delta = \frac{\varepsilon}{|m|} \). This guarantees that \( |m||x - c| < \varepsilon \) whenever \( 0 < |x - c| < \delta \).
Epsilon-Delta Definition
The epsilon-delta definition is foundational in calculus as it rigorizes the concept of limits. It offers a way to formally declare that a function is approaching a particular value as the input approaches a specific point. This method clarifies what it means for the output of a function to be "near" a given value.
In essence, the epsilon-delta definition states that for each \( \varepsilon > 0 \) (no matter how tiny), we can find a \( \delta > 0 \) such that whenever the input \( x \) is within \( \delta \) units of a particular point \( c \) (but not equal to \( c \)), the output \( f(x) \) stays within \( \varepsilon \) units of \( L \). This idea helps in crafting proofs in calculus by solidifying how limits operate.
In essence, the epsilon-delta definition states that for each \( \varepsilon > 0 \) (no matter how tiny), we can find a \( \delta > 0 \) such that whenever the input \( x \) is within \( \delta \) units of a particular point \( c \) (but not equal to \( c \)), the output \( f(x) \) stays within \( \varepsilon \) units of \( L \). This idea helps in crafting proofs in calculus by solidifying how limits operate.
- \( \varepsilon \) sets how close the function's value must be to the limit.
- \( \delta \) dictates how close \( x \) needs to be to \( c \).
Other exercises in this chapter
Problem 20
In Problems 1-32, use a table or a graph to investigate each limit. $$ \lim _{x \rightarrow 0} \frac{x}{1-e^{-x}} $$
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(a) Use a graphing calculator to sketch the graph of $$f(x)=e^{a x} \sin x, \quad x \geq 0$$ for \(a=-0.1,-0.01,0,0.01\), and 0.1. (b) Which part of the functio
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In Problems 15-24, find the values of \(x \in\) R for which the given functions are both defined and continuous. $$ f(x)=\frac{x}{x+1} $$
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Evaluate the limits. $$ \lim _{x \rightarrow \infty} \frac{3}{2+e^{-x}} $$
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