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The epsilon-delta definition is a formal way to understand limits in calculus. It's a method to prove that the limit of a function is a specific value. This definition plays a crucial role in establishing the precise behavior of functions as they approach a particular point or as they extend towards infinity. In simple terms, the definition states that for every small, positive number, \( \epsilon \), you can find a corresponding value, \( N \) (or \( \delta \) in some contexts), such that whenever the input, \( x \), is beyond this value, the output, \( f(x) \), remains within \( \epsilon \) distance from the limit \( L \). This is often expressed as:

• If \( x \to \infty \): For every \( \epsilon > 0 \), there exists an \( N \) such that if \( x > N \), then \( |f(x) - L| < \epsilon \).
• If \( x \to -\infty \): For every \( \epsilon > 0 \), there is an \( N \) so that if \( x < N \), then \( |f(x) - L| < \epsilon \).

In the given exercise, \( \epsilon = 0.001 \) and the limit \( L \) is 1 as \( x \to -\infty \). This means we are tasked to find an \( N \) such that for any \( x < N \), the function \( \frac{x}{x+1} \) stays within 0.001 of the value 1.

Solving inequalities in calculus involves finding the range of values that satisfy a given condition in terms of \( x \). In the problem, the inequality is \( \left| \frac{x}{x+1} - 1 \right| < 0.001 \). To solve this:

• Simplify the expression inside the absolute value. Here, the difference \( \frac{x}{x+1} - 1 \) simplifies to \( \frac{-1}{x+1} \).
• Re-evaluate the inequality without absolutes: \( \frac{1}{|x+1|} < 0.001 \).
• This can be rearranged to \( |x+1| > 1000 \).

Since \( x < -1 \) in this context, the inequality translates to \( x+1 < -1000 \), leading to \( x < -1001 \). This step helps us identify that all values less than -1001 will meet the condition. Hence, solving inequalities like this transforms the problem into manageable steps, ensuring it meets the specific \( \epsilon \) criteria.

The limit of a function describes the behavior of \( f(x) \) as \( x \) approaches a specific point or infinity. In simpler terms, it helps us understand the value that \( f(x) \) gets closer to as \( x \) becomes very large or very small. For this exercise, we are interested in the limit as \( x \to -\infty \). This tells us what value \( \frac{x}{x+1} \) approaches as \( x \) goes to negative infinity.

• The general concept of a limit involves envisioning the output of a function hovering around a particular value as the input zooms off to infinity or approaches a particular number.
• Here, the function \( \frac{x}{x+1} \) simplifies under the lens of limits: as \( x \to -\infty \), both the numerator and the denominator behave predominantly like \( x \), so they nearly cancel.
• This results in the function value approaching 1 as \( x \to -\infty \), which is calculated and validated by substituting values, resulting in a satisfactory approximation quite close to that of 1.

The concept of a limit underpins many foundational ideas in analysis and calculus by formalizing how functions behave near points of interest or at infinity.