91Ó°ÊÓ

When dealing with limits, we want to understand the behavior of a function as the input values get increasingly close to a particular point. In the context of continuity, this often involves finding the limit of a function as it approaches a given point, represented as \( \lim_{x \to a} f(x) \).
The process requires examining how the function behaves near that point. We check if the function approaches a specific value as \( x \) gets closer to \( a \).

• If the limit exists and is finite, it means as \( x \) hovers around \( a \), \( f(x) \) gets closer and closer to some number.
• If the limit doesn't exist, the function behaves unpredictably or shoots off to infinity.

In our exercise, we calculated \( \lim_{x \to 5} \frac{x-5}{x+5} \). Since both the numerator and denominator become simple numbers (0 and 10), we could find the limit, which turned out to be 0. This was helpful for confirming continuity at a point.

Function evaluation is the process of finding the value of a function at a specific point. It involves substituting the given number for the variable in the function's formula. This step tells us whether the function is defined at the point we're interested in.
For instance, with our exercise's function \( f(x) = \frac{x-5}{x+5} \) and point \( a = 5 \), inserting 5 into the function helps determine \( f(a) \).

• This is done by substituting: \( f(5) = \frac{5-5}{5+5} \).
• As \( 5-5 = 0 \) and \( 5+5 = 10 \), \( f(5) \) simplifies to \( \frac{0}{10} = 0 \).

Therefore, the function \( f \) is defined and equals 0 at \( x = 5 \). The existence of \( f(a) \) is a piece of the puzzle in determining continuity.

Direct substitution is a simple and often-used method for finding limits and evaluating functions. It involves plugging the point of interest directly into the function.
If both numerator and denominator are non-zero, direct substitution can often efficiently yield the limit or value of the function. In our example, we performed direct substitution to find both \( f(a) \) and the limit as \( x \) approaches 5.

• The function \( f(x) = \frac{x-5}{x+5} \) was evaluated by plugging 5 directly into the function.
• This yielded \( \frac{5-5}{5+5} = \frac{0}{10} = 0 \).

Substituting directly helped confirm the limit also equals the function's value at \( x = 5 \). When these two match, they confirm the continuity of the function at that point without further complex calculations.