91Ó°ÊÓ

The epsilon-delta definition of a limit provides a formal foundation for understanding limits in calculus. This definition revolves around two critical components, \( \varepsilon \) (epsilon) and \( \delta \) (delta).

• \( \varepsilon \), a positive number, indicates how close the value of the function needs to be to the limit \( L \).
• \( \delta \), another positive number, shows how close \( x \) needs to be to \( c \) to ensure that the function's value remains within \( \varepsilon \) of \( L \).

The goal is to find a \( \delta \) such that for every \( x \) satisfying \( 0 < |x - c| < \delta \), the inequality \( |f(x) - L| < \varepsilon \) holds true.
This rigorous approach helps verify the limit \( L \) by ensuring that the function \( f(x) \) behaves predictably as \( x \) approaches \( c \). In our exercise, we need \( \delta \) to make sure the simplified function \( f(x) = x + 1 \) stays close to \( L = -4 \) with \( \varepsilon = 0.05 \).

Evaluating the limit of a function involves simplifying it and determining its value as \( x \) approaches a particular point, \( c \).
In the given exercise, the function \( f(x) = \frac{x^2 + 6x + 5}{x + 5} \) needs simplification before finding the limit.

• First, factorize the numerator: \( x^2 + 6x + 5 = (x + 5)(x + 1) \).
• Cancel the common term \( (x + 5) \) found in both the numerator and denominator, simplifying it to \( f(x) = x + 1 \) for \( x eq -5 \).

With this simplified function, evaluate \( \lim_{x \to -5} f(x) \):
- Substitute \( x = -5 \) into the simplified \( f(x) = x + 1 \).
- The result is \( -5 + 1 = -4 \), thus \( L = -4 \).
The limit evaluation confirms that \( f(x) \) approaches \( -4 \) as \( x \) gets closer to \( -5 \), allowing us to move on to the epsilon-delta condition.

Rational functions, like the one in our exercise, are expressions formed by the ratio of two polynomials. They are written in the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials.
Key characteristics include:

• They can be simplified by factoring and canceling common terms, as demonstrated in the simplification of our function.
• Care must be taken around points where \( Q(x) = 0 \) to identify and handle any discontinuities or holes in the function.

In the original problem, \( f(x) = \frac{x^2 + 6x + 5}{x + 5} \) is initially a rational function. Simplified to \( f(x) = x + 1 \) after factoring and canceling \( (x + 5) \), it no longer has the rational expression form at \( x eq -5 \).
Rational functions often need simplification to apply the epsilon-delta definition or to evaluate limits efficiently. Simplifying \( f(x) \) allowed the limit to be easily calculated at \( c = -5 \), illustrating why understanding rational functions is a crucial skill in calculus.
By recognizing and simplifying, we contributed to solving our problem and verifying our limit.