Problem 29 - Calculus: Early Transcendentals

Question

Limit proofs Use the precise definition of a limit to prove the following limits. Specify a relationship between \(\varepsilon\) and \(\delta\) that guarantees the limit exists. $$\lim _{x \rightarrow 2}\left(x^{2}+3 x\right)=10$$

Step-by-Step Solution

Answer

Question: Prove that the limit of the function (x^2 + 3x) as x approaches 2 is equal to 10, using the precise (ε-δ) definition of a limit. Short Answer: The limit of the function (x^2 + 3x) as x approaches 2 is equal to 10 because, for any ε > 0, we can find a δ > 0 (given by δ = min{sqrt(ε), 1}) that makes |(x^2 + 3x) - 10| < ε when 0 < |x - 2| < δ.

Step 1: Write the precise (ε-δ) definition of a limit

Recall that for a function f(x), the limit as x approaches a (in this case 2) is L (in this case 10) if, for any ε > 0, there exists a δ > 0 such that if 0 < |x - 2| < δ, then |f(x) - 10| < ε. That is, $$\forall \varepsilon>0, \exists \delta>0 : 0<|x-2|<\delta \Rightarrow |(x^{2}+3x)-10|<\varepsilon$$

Step 2: Substitute the function and the limit into the inequality

Replace f(x) with the given function, x^2 + 3x, and L with the given limit, 10. Then, the definition becomes: $$0<|x-2|<\delta \Rightarrow |(x^{2}+3x)-10|<\varepsilon$$

Step 3: Simplify the expression

First, simplify the expression (x^2 + 3x) - 10: \begin{align*} (x^{2}+3x)-10 &= x^{2} + 3x - 10 \\ &= x^2 + 3x - 4 - 6 \\ &= (x^2 - 4) + 3x - 6 \\ &= (x - 2)(x + 2) + 3x - 6 \end{align*} Then the inequality becomes: $$0<|x-2|<\delta \Rightarrow |(x - 2)(x + 2) + 3x - 6|<\varepsilon$$

Step 4: Find a relationship between ε and δ

We need to find a δ such that the inequality is true for all ε > 0. We can use the fact that |x - 2| < δ: \begin{align*} |(x - 2)(x + 2) + 3x - 6| &= |x - 2||x + 2 + 3| \\ &= |x - 2||x + 5| \end{align*} Since |x - 2| < δ, we can bound |x + 5| using the fact that |x - 2 + 7| = |x + 5|: $$|x + 5| \le |x - 2| + 7 < \delta + 7$$ So, |(x - 2)(x + 2) + 3x - 6| = |x - 2||x + 5| < δ(|x+5|): $$|(x - 2)(x + 2) + 3x - 6| < \delta(|x+5|) < \delta(\delta + 7)$$ To ensure the inequality in step 3, we can set: $$\varepsilon = \delta(\delta + 7)$$ Then, we can choose δ to be the minimum of the square roots of ε and 1, that is: $$\delta = \min\{\sqrt{\varepsilon}, 1\}$$

Step 5: Conclude the proof

In summary, we showed that for any ε > 0, there exists a δ > 0 such that if 0 < |x - 2| < δ, then |(x^2 + 3x) - 10| < ε, given by the relation: $$\delta = \min\{\sqrt{\varepsilon}, 1\}$$ Therefore, by the precise (ε-δ) definition of a limit: $$\lim _{x \rightarrow 2}\left(x^{2}+3 x\right)=10$$

Key Concepts

epsilon-delta definition

The epsilon-delta (\varepsilon-\delta) definition is a concept in calculus used to precisely define the limit of a function. Imagine you have a function \( f(x) \) and you're interested in what happens to \( f(x) \) as \( x \) gets closer and closer to some number \( a \). The epsilon-delta definition provides a way to express that as \( x \) approaches \( a \), the function \( f(x) \) approaches a specific value \( L \).

• \( \varepsilon \) (epsilon) represents any small positive number, indicating how close \( f(x) \) should be to \( L \).
• \( \delta \) (delta) depends on the chosen \( \varepsilon \) and tells us how close \( x \) needs to be to \( a \).

For every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - a| < \delta \), then \( |f(x) - L| < \varepsilon \). This means: to make \( f(x) \) as close as desired to \( L \), \( x \) just needs to stay sufficiently close to \( a \). This formalism allows for rigorous proofs of limits.

precise definition of limit

The precise definition of a limit centers around making the concept of approaching a value as exact as possible. For a function \( f(x) \), saying the limit as \( x \) approaches a value \( a \) is \( L \) means that no matter how tight the band around \( L \) is (determined by \( \varepsilon \)), we can find an appropriate range (determined by \( \delta \)) for \( x \). This ensures \( f(x) \) stays within the \( \varepsilon \)-distance of \( L \).
For instance, if considering \( \lim_{x \to 2} (x^2 + 3x) = 10 \), we want to show that \( (x^2 + 3x) \) is as near to 10 as desired when \( x \) is near 2. Here,

• \( a = 2 \)
• \( L = 10 \)

By selecting any \( \varepsilon \), you find a corresponding \( \delta \) so that when \( x \) is in the interval \( (2 - \delta, 2 + \delta) \), \( f(x) \) will remain confined within \( (10 - \varepsilon, 10 + \varepsilon) \). This quantitative approach is what makes limits reliable and rigorously proven in calculus.

relationship between epsilon and delta

An important aspect of using the epsilon-delta definition is establishing a precise relationship between \( \varepsilon \) and \( \delta \). This relationship is crucial for proving that a limit exists. Essentially, for any given \( \varepsilon \), a corresponding \( \delta \) is determined so that the formal criteria for limits are satisfied. This connection ensures the output \( f(x) \) is adequately close to the limit \( L \).
In the problem \( \lim_{x \to 2} (x^2 + 3x) = 10 \), for every chosen \( \varepsilon \), a calculation yields a \( \delta = \min\{\sqrt{\varepsilon}, 1\} \) such that if \( 0 < |x - 2| < \delta \), then \( |(x^2 + 3x) - 10| < \varepsilon \). This relationship confirms that as \( \varepsilon \) becomes smaller, the chosen \( \delta \) gets appropriately closer to ensure accuracy of the limit. - Ensuring \( \varepsilon \) and \( \delta \) are correctly related ensures the precision needed for confirming limits.- This relationship guarantees that the limit value is consistently approachable by the function \( f(x) \) as \( x \) is drawn as close as desired to a certain number.
Finding this relationship mathematically can require algebraic manipulation and sometimes creative reassignment based on the function's behavior around the point \( a \). This maintains the precise connection between variations in inputs and outputs.