91Ó°ÊÓ

In calculus, limits play a crucial role in understanding how functions behave as they approach specific values. Specifically, a limit tells us the value that a function approaches as the input comes closer to a particular point.

For example, consider the function provided in the exercise: \[\frac{x^2 - 9}{x + 3}\]When trying to find the limit as \( x \) approaches -3, we encounter an undefined form, 0/0, which means it's necessary to simplify the expression to determine the limit.

• Start by factoring the numerator, \( x^2 - 9 \), since it's a difference of squares. This results in \((x + 3)(x - 3)\).
• Cancel the \( (x + 3) \) term from both the numerator and the denominator for \( x eq -3 \).
• The function simplifies to \( x - 3 \), allowing us to easily find the limit by substituting \( x = -3 \).

Therefore, the limit \( \lim_{x \to -3} \frac{x^2 - 9}{x + 3} \) results in \( -6 \). Assigning this to \( k \), we seamlessly connect the function's behavior at this point with its limit.

Continuity is about ensuring a function has no abrupt changes, gaps, or breaks in its graph at a point. A function is continuous at a point if the following conditions are satisfied:

• The function is defined at the point.
• The limit of the function exists as the input approaches the point.
• The limit equals the actual value of the function at that point.

In the context of this exercise, we're investigating the point \( x = -3 \) to determine if \( f(x) \) is continuous there. Originally, \( f(x) = \frac{x^2 - 9}{x + 3} \) is not defined at \( x = -3 \), due to division by zero. However, by assigning \( k = -6 \) based on the limit, the function becomes continuous:

• At \( x = -3 \), define \( f(x) = k = -6 \).
• Both the limit as \( x \to -3 \) and \( f(-3) \) take the value of \(-6\), fulfilling the continuity conditions.

Factoring polynomials is a powerful algebraic tool for simplifying expressions, solving equations, and finding limits. Essentially, factoring involves breaking down a polynomial into simpler, multiplied components.

In our function, we started with:
\[\frac{x^2 - 9}{x + 3}\]

By recognizing \( x^2 - 9 \) as a difference of squares, we can factor it as follows:
\[x^2 - 9 = (x + 3)(x - 3)\]
This allows us to cancel out the common \( (x + 3) \) term in both the numerator and the denominator, simplifying the function significantly.

• For terms like \( x^2 - b^2 \), use the identity \( x^2 - b^2 = (x - b)(x + b) \).
• Look for common factors in both the numerator and the denominator to simplify.

By transforming the expression to \( x - 3 \), we can handle limits and continuity more effectively, making it a consistent polynomial function everywhere except the initially undefined \( x = -3 \).