# Article: Understanding the Limit of Algebraic Functions
In calculus, the concept of a limit is fundamental to the study of change and motion. It allows us to comprehend how functions behave as inputs approach a certain value. This foundational idea is especially pertinent when examining algebraic functions, which are built from polynomials and rational expressions.
Algebraic functions are those that can be expressed using algebraic operations such as addition, subtraction, multiplication, division, and raising to a rational power.
## Definition of a Limit
The formal definition of a limit goes as follows: Let \(f(x)\) be a function defined near the point \(a\), except possibly at \(a\) itself. We say that the limit of \(f(x)\) as \(x\) approaches \(a\) is \(L\), and write it as:
\[
\lim_{x \to a} f(x) = L
\]
if for every number \(\epsilon > 0\), there is a number \(\delta > 0\) such that if \(0 < |x - a|< \delta\), then \(|f(x) - L|< \epsilon\). ## Finding Limits in Algebraic Functions To find the limit of an algebraic function, you often employ direct substitution. However, there are cases where direct substitution results in forms that are not immediately resolvable, such as \(0/0\). In such cases, algebraic manipulation techniques such as factoring, conjugates, or common denominators are used to simplify the function for a limit to be determined. ### Indeterminate Forms An algebraic function may lead to the following indeterminate forms when evaluating a limit:
- \(0/0\) - \(\infty/\infty\) - \(0 \times \infty\) - \(\infty - \infty\) - \(0^0\) - \(\infty^0\) - \(1^\infty\) Each of these forms requires a particular strategy or algebraic manipulation to determine the limit. ## Problems and Solutions Here are 20 problems related to limits of algebraic functions, along with their solutions: ### Problem 1 Evaluate the limit: \[ \lim_{x \to 2} (x^2 + 3x - 10) \] **Solution:** By direct substitution: \[ \lim_{x \to 2} (x^2 + 3x - 10) = 2^2 + 3(2) - 10 = 4 + 6 - 10 = 0 \] ### Problem 2 Determine the limit: \[ \lim_{x \to -1} \frac{x^2 - 1}{x + 1} \] **Solution:** Factor and simplify: \[ \lim_{x \to -1} \frac{(x + 1)(x - 1)}{x + 1} = \lim_{x \to -1} (x - 1) = -1 - 1 = -2 \] ### Problem 3 Find the limit: \[ \lim_{x \to 4} \frac{2x - 8}{x^2 - 16} \] **Solution:** Factor and reduce: \[ \lim_{x \to 4} \frac{2(x - 4)}{(x + 4)(x - 4)} = \lim_{x \to 4} \frac{2}{x + 4} = \frac{2}{4 + 4} = \frac{1}{4} \] ### Problem 4 Calculate the limit: \[ \lim_{x \to 3} (5x^3 - 4x + 7) \] **Solution:** By direct substitution: \[ \lim_{x \to 3} (5x^3 - 4x + 7) = 5(3)^3 - 4(3) + 7 = 135 - 12 + 7 = 130 \]
