91Ó°ÊÓ

The Theorem on Limits of Rational Functions is an essential concept when dealing with limits involving rational functions. According to this theorem, if you have a rational function, say \( R(x) = \frac{P(x)}{Q(x)} \), and you want to find the limit as \( x \) approaches a value \( c \), you can use direct substitution if \( Q(c) eq 0 \). This works because rational functions, like other functions, are defined and continuous at points where the denominator is not zero.
For example, if we have a rational function \( R(x) = \frac{x^2 + 3x}{x-2} \) and need to find \( \lim_{x \rightarrow 1} R(x) \), we perform direct substitution: \( R(1) = \frac{1^2 + 3 \times 1}{1 - 2} = \frac{4}{-1} = -4 \).
However, if the denominator were zero at \( x = 1 \), you could not directly substitute the value and would need to analyze the behavior around that point to determine if the limit exists.

Continuous functions are fundamental in calculus, especially when working with limits. A function is continuous at a point \( c \) if the limit as \( x \) approaches \( c \) is equal to the function's value at \( c \): \( \lim_{x \rightarrow c} f(x) = f(c) \). Polynomial functions, such as the one in our example, \( f(x) = x^2 + 3 \), are especially nice because they are continuous everywhere.
This means that for any value you pick to substitute into a polynomial, the limit can be found easily by evaluating the function directly at that point. There are no jumps, holes, or asymptotes interfering with the computation. This property simplifies the process of finding limits dramatically for polynomial and other continuous functions.

Direct Substitution is the technique used to find the limit of a function by replacing the variable with the value that it approaches. This method is straightforward and very efficient when the function in question is continuous at the point of interest.
In our problem, we wanted to find \( \lim_{x \rightarrow -2} \( x^2 + 3 \) \). Since \( x^2 + 3 \) is a polynomial (and thus continuous everywhere), we can use direct substitution:

• Substitute \( x = -2 \) into the function: \( f(-2) = (-2)^2 + 3 \).
• Calculate: \( (-2)^2 + 3 = 4 + 3 = 7 \).

This confirms that \( \lim_{x \rightarrow -2} \( x^2 + 3 \)\). By utilizing direct substitution, we see that as \( x \) approaches \( -2 \), the value of the function approaches 7. Direct substitution is a powerful tool when the function's continuity ensures that the limit process is seamlessly straightforward.