Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Give an \(\varepsilon-\delta\) proof of each limit fact. $$ \lim _{x \rightarrow-1}\left(x^{2}-2 x-1\right)=2 $$

Step-by-Step Solution

Answer

Choose \(\delta = \min(1, \frac{\varepsilon}{5})\) for the limit proof.

Step 1: Understanding the Limit Statement

The limit statement \[\lim _{x \rightarrow-1}\left(x^{2}-2 x-1\right)=2\]states that as \(x\) approaches \(-1\), the value of \(f(x) = x^2 - 2x - 1\) gets arbitrarily close to \(2\). We need to prove this formally using the \(\varepsilon-\delta\) definition.

Step 2: Recall the Definition of Limit

The \(\varepsilon-\delta\) definition of a limit states that for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that whenever \(0 < |x + 1| < \delta\), it follows that \(|x^2 - 2x - 1 - 2| < \varepsilon\).

Step 3: Simplify the Expression

First, simplify the expression \[|x^2 - 2x - 1 - 2| = |x^2 - 2x - 3|.\]We are required to find a \(\delta\) for every \(\varepsilon\) such that when \(0 < |x + 1| < \delta\), \(|x^2 - 2x - 3| < \varepsilon\) holds true.

Step 4: Factor and Further Simplify

Factor the expression \[x^2 - 2x - 3 = (x - 3)(x + 1).\]Hence, \[|x^2 - 2x - 3| = |(x - 3)(x + 1)|.\]

Step 5: Relate to \(\varepsilon\)

We want \[|(x - 3)(x + 1)| < \varepsilon.\]Assume \(|x + 1| < 1\), implying \(-2 < x < 0\). This keeps \(x\) in a bounded interval.

Step 6: Bound the Expression \(|x - 3|\)

Within the interval \(-2 < x < 0\), \(|x - 3|\) is \[|x - 3| < 5.\]This is because the maximum difference within this interval is 5 (e.g., at \(x = 0\)).

Step 7: Choose \(\delta\) Appropriately

Using the bound, \[|(x - 3)(x + 1)| < 5|x + 1| < \varepsilon.\]To satisfy this inequality, choose \[|x + 1| < \min\left(1, \frac{\varepsilon}{5}\right),\]ensuring \(|x^2 - 2x - 3| < \varepsilon\).

Step 8: Conclusion of \(\varepsilon-\delta\) Proof

Thus, for every \(\varepsilon > 0\), choosing \[\delta = \min\left(1, \frac{\varepsilon}{5}\right)\]ensures that if \(0 < |x + 1| < \delta\), then \(|x^2 - 2x - 1 - 2| < \varepsilon\). Hence, \[\lim _{x \rightarrow-1}\left(x^{2}-2 x-1\right)=2.\]

Key Concepts

Limit of a Function

The concept of a limit of a function is fundamental in calculus. Essentially, it describes the behavior of a function as the input approaches a certain value. In our case, we're interested in what happens to the function \( f(x) = x^2 - 2x - 1 \) as \( x \) approaches \( -1 \). This is written as \( \lim_{x \rightarrow -1} (x^2 - 2x - 1) = 2 \). It tells us that as \( x \) gets closer and closer to \( -1 \), the function value gets closer to \( 2 \). This doesn't mean the function value is exactly \( 2 \) at \( x = -1 \), but rather it is very close when \( x \) is very near \( -1 \).

If you think of the limit as a road sign showing how the direction you're headed points toward a particular destination (here, \( 2 \)), even if you never actually reach exactly \( x = -1 \), the function's value trends toward \( 2 \).

• It helps in studying the behavior of functions near specific points.
• Not always about finding the exact value at the point but understanding the trend.
• Used extensively in calculus to discuss continuity and differentiation.

Epsilon-Delta Definition

The \( \varepsilon-\delta \) definition is a formal way to define what it means for a function to have a limit at a point. It's the rigorous backbone of calculus limits, which avoids any ambiguity. Here's a summary:

• Given any number \( \varepsilon > 0 \), there should be a number \( \delta > 0 \) such that for all \( x \) satisfying \( 0 < |x + 1| < \delta \), the expression \(|f(x) - L| < \varepsilon \) holds true.
• In simpler words, no matter how tiny \( \varepsilon \) is, there exists a range for \( x \) (controlled by \( \delta \)) that ensures the function value stays within this "target" range around the limit \( L \).

This gives us a precise mathematical way to express the closeness of \( f(x) \) to \( L \) as \( x \) approaches the target point (like \(-1\) in our exercise).

This approach turns "close to" into something concrete. It tells us exactly how close \( x \) needs to be to \(-1\), and subsequently how close \( f(x) \) will be to \( 2 \), using a logical and repeatable method.

Polynomial Limits

Calculating limits for polynomials, such as \( x^2 - 2x - 1 \), can often be simpler than more complex functions due to the continuous nature of polynomials.

• Polynomials are smooth and have no breaks, holes, or cusps.
• They are defined for all real numbers, making limit calculations straightforward.

For polynomials, if you need the limit as \( x \) approaches some value, you can usually substitute the value directly into the polynomial (provided no division by zero occurs).

However, in an \( \varepsilon-\delta \) context, as in our example, we need to confirm that as \( x \) gets close to a point like \(-1\), the polynomial behaves predictably, approaching the expected limit. This involves conserving a sensible range for \( x \) and its image \( f(x) \) to guarantee equations like \(|(x-3)(x+1)| < \varepsilon \) can be satisfied.

• This often requires manipulating the polynomial into a factored form to simplify calculating bounds.

By factoring \( x^2 - 2x - 3 \) to \( (x-3)(x+1) \), it aligns the operation with the \( \varepsilon-\delta \) technique, demonstrating that polynomial limits can be systematically transformed into standards like these to ensure accuracy and understanding.