Problem 23 - Thomas Calculus

Question

Limits of quotients Find the limits in Exercises \(23-42\) $$\lim _{x \rightarrow 5} \frac{x-5}{x^{2}-25}$$

Step-by-Step Solution

Answer

The limit is \(\frac{1}{10}\).

Step 1: Identify the form of the limit

First, substitute \(x = 5\) into the expression \(\frac{x-5}{x^2-25}\). This gives \(\frac{0}{0}\), an indeterminate form. Therefore, further algebraic manipulation is needed to resolve the limit.

Step 2: Factor the denominator

Notice that the denominator \(x^2-25\) is a difference of squares. Recall the identity \(a^2-b^2 = (a-b)(a+b)\), which can be applied to factorize \(x^2-25\) as \((x-5)(x+5)\).

Step 3: Simplify the expression

After factoring, the expression becomes \(\frac{x-5}{(x-5)(x+5)}\). Notice that \(x-5\) appears in both the numerator and denominator. As \(x\rightarrow 5\), we can cancel out \(x-5\) from both, leading to a simplified form \(\frac{1}{x+5}\).

Step 4: Evaluate the limit

Now substitute \(x = 5\) back into the simplified expression \(\frac{1}{x+5}\). This gives \(\frac{1}{5+5} = \frac{1}{10}\). The limit is thus \(\frac{1}{10}\).

Key Concepts

Indeterminate Forms

In calculus, evaluating limits often involves encountering expressions like \(\frac{0}{0}\), known as indeterminate forms. These are not directly solvable because both the numerator and denominator tend to zero simultaneously. Indeterminate forms indicate that more algebraic manipulation is required to simplify the expression without losing the value of the overall limit. Such manipulations typically include factoring or applying techniques to rewrite the expression. This process becomes essential especially in cases involving polynomials or rational functions, as it allows us to find a clear and more simplified version of the function to evaluate the limit correctly.

Factoring Polynomials

Factoring polynomials is a key technique used to simplify algebraic expressions and solve limits. Consider the difference of squares, a common expression found in limits problems, represented by the formula \(a^2 - b^2 = (a-b)(a+b)\). In our case, when encountering the polynomial \(x^2 - 25\), we recognize it as a difference of squares where \(a = x\) and \(b = 5\). Thus, it can be factored into \((x-5)(x+5)\). This factorization is critical because it often allows you to cancel out terms in the numerator and denominator, facilitating the simplification of the expression. By factoring, we prepare the expression for straightforward evaluation at the limit point, especially when initially presented with an indeterminate form.

Simplifying Algebraic Expressions

Once factors have been identified, the next step is simplifying the algebraic expression. This often involves canceling out common terms from the numerator and the denominator. For instance, in the given problem, after factoring \(x^2 - 25\) as \((x-5)(x+5)\), you notice \(x-5\) appears in both the numerator and denominator. This common factor can be canceled, resulting in the simplified expression \(\frac{1}{x+5}\). Simplification is vital as it moves the expression away from an indeterminate form. With the expression simplified, evaluating the limit becomes straightforward, as substitution no longer results in an undefined operation. It's a crucial skill in analyzing and resolving limits and ensures accurate calculation of their values.