Question

Prove the given limit using an \(\varepsilon-\delta\) proof. $$ \lim _{x \rightarrow 5}(3-x)=-2 $$

Short answer

The limit is proven with \( \delta = \varepsilon \) using the \( \varepsilon-\delta \) definition.

Step-by-step solution

Understand the Definition

The formal definition of a limit is: \( \lim_{x \to a} f(x) = L \) if for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( 0 < |x - a| < \delta \), it follows that \( |f(x) - L| < \varepsilon \). In this problem, we have \( a = 5 \), \( f(x) = 3-x \), and \( L = -2 \).

Express the Function and Limit Conditions

We want to show that for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - 5| < \delta \), then \( |(3 - x) + 2| < \varepsilon \). Simplifying the absolute value, we get \( |x - 5| < \varepsilon \).

Choose an Appropriate Delta

From the expression \( |x - 5| < \varepsilon \), we can directly choose \( \delta = \varepsilon \). This choice of \( \delta \) ensures that whenever \( 0 < |x - 5| < \delta \), the condition \( |(3 - x) + 2| < \varepsilon \) is satisfied.

Show the Validity of the Delta Choice

With \( \delta = \varepsilon \), if \( 0 < |x - 5| < \delta \), then it directly implies \( |(3 - x) + 2| = |x - 5| < \varepsilon \). This satisfies the condition that \( |f(x) - L| < \varepsilon \), hence proving the limit.

Conclusion

We have shown that for every \( \varepsilon > 0 \), choosing \( \delta = \varepsilon \) works to satisfy the \( \varepsilon-\delta \) definition for the limit. Therefore, \( \lim_{x \to 5}(3-x) = -2 \).

Key concepts

Limit Definition

The concept of limits in calculus involves understanding how a function behaves as the input approaches a certain point. The formal epsilon-delta definition of a limit is crucial for rigorous proofs. It states that for a function \( f(x) \), if \( \lim_{x \to a} f(x) = L \), then for every positive number \( \varepsilon \), no matter how small, there exists a positive number \( \delta \) such that if the input \( x \) is within \( \delta \) units of \( a \) (but not equal to \( a \)), then the function \( f(x) \) is within \( \varepsilon \) units of \( L \).

This definition allows us to precisely capture what it means for a function to approach a particular value as the input nears a specified point. It sets the foundation for many other concepts in calculus, including derivatives and integrals, making it essential for every student to grasp.

Delta Choice

Choosing the right \( \delta \) is an essential part of creating a valid epsilon-delta proof. The goal is to find a \( \delta \) that works for every possible \( \varepsilon \), ensuring that the function value \( f(x) \) stays within \( \varepsilon \) of \( L \).

In our specific problem, with the function \( f(x) = 3 - x \) and limit \( \lim_{x \to 5} (3-x) = -2 \), the task is simplified because setting \( |x - 5| < \varepsilon \) neatly fits the requirement. Thus, our choice of \( \delta = \varepsilon \) works perfectly. This makes the proof more straightforward as the calculation directly satisfies the given limit condition, showing that selecting the "right" \( \delta \) is often the key to a successful limit proof.

Absolute Value Simplification

An important technique in epsilon-delta proofs involves simplifying expressions using absolute value properties. The absolute value function, \( |x| \), gives the non-negative distance from zero, and simplifies many expressions in calculus.

In this exercise, we have the expression \( |(3 - x) + 2| \), which simplifies to \( |x - 5| \). This step reduces the complexity of the argument by linking it directly to the statement \( |x - a| < \delta \).

Understanding how to simplify using absolute values helps in managing expressions more effectively. It ensures that the focus remains on the main goal: proving the limit conditions by capturing all behaviors towards approaching that limit.

Continuity

Continuity is closely related to limits, providing insight into the behavior of functions without abrupt changes. A function is continuous at a point \( a \) if \( \lim_{x \to a} f(x) = f(a) \).

In simpler terms, a function is continuous if it can be drawn without lifting the pen. For the example of \( f(x) = 3-x \), it is continuous everywhere because it is a linear function.

By understanding continuity, we can see how epsilon-delta proofs confirm continuous behaviors at specific points. This is because if a limit exists at \( a \) and equals the function value, the function at that point is smooth and predictable.