Problem 5 - Technical Mathematics with Calculus

Question

Evaluate each limit. $$\lim _{x \rightarrow 5} \frac{x^{2}-25}{x-5}$$

Step-by-Step Solution

Answer

The limit evaluates to 10.

Step 1: Factor the numerator

Recognize that the numerator is a difference of squares, which can be factored. The factoring formula is \(a^{2} - b^{2} = (a + b)(a - b)\). Apply this to factor the numerator: \(x^{2} - 25 = (x - 5)(x + 5)\).

Step 2: Simplify the expression

After factoring the numerator, the expression can be simplified by canceling out the common \(x - 5\) terms in the numerator and denominator. The simplified expression will be \(\frac{(x + 5)}{1}\) which is just \(x + 5\).

Step 3: Evaluate the limit

Now that the expression is simplified, substitute the value of \(x\) that the limit approaches (in this case \(x = 5\)) into the simplified expression to evaluate the limit: \(5 + 5 = 10\).

Key Concepts

Difference of Squares

One of the fundamental algebraic patterns used in calculus is the difference of squares. It is a specific case of factoring polynomials that can transform complex expressions into simpler ones. The formula for the difference of squares is \( a^2 - b^2 = (a + b)(a - b) \).

This is particularly useful when dealing with limits involving quadratic expressions, as it allows us to break down expressions that are difficult to evaluate directly. If you come across a term like \( x^2 - 25 \), you can identify \( a = x \) and \( b = 5 \), which fits the difference of squares form, allowing you to factor it as \( (x + 5)(x - 5) \) before proceeding with further simplification.

Factoring Polynomials

When evaluating limits in calculus, it is quite common to encounter polynomials that can be simplified through factoring. Factoring a polynomial involves rewriting it as a product of its factors. This can simplify complex expressions and allow us to cancel out terms to make limit computation easier. For example, in the provided exercise, the expression \( x^2 - 25 \) was factored into \( (x + 5)(x - 5) \).

This approach is crucial when an expression appears to be undefined or indeterminate in its current form. By factoring, we can often remove the part of the expression that is causing the indeterminate form and proceed with finding the limit.

Simplifying Expressions

Simplifying expressions is essential for solving calculus problems effectively. It involves reducing an expression into its most basic form without changing its value. Simplification can include factoring polynomials, canceling out terms, combining like terms, and other algebraic manipulations. In the context of limits, simplifying allows us to get rid of the terms that complicate direct substitution. In our exercise, after factoring, we cancelled the common term \( x - 5 \) in the numerator and denominator, which simplified the expression to \( x + 5 \) and made it possible to directly substitute \( x = 5 \) to find the limit.

Limits in Calculus

The concept of a limit is at the heart of calculus. Specifically, it describes the behavior of a function as the input (or 'x' value) approaches a certain point. Limits can show us the value that a function approaches, even if it doesn't actually reach that value at the point in question. In our exercise, we evaluate \( \lim_{x \rightarrow 5} \frac{x^{2}-25}{x-5} \), which involves understanding the behavior of this function as \( x \) approaches 5.

To compute it, we first simplified the expression using the earlier discussed methods and then substituted the limit point \( x = 5 \) into the simplified expression. By mastering the process of evaluating limits we build a foundational skill used throughout calculus, from derivatives to integrals.