Problem 50 - Calculus of a Single Variable

Question

Finding a Limit In Exercises \(47-62,\) find the limit. $$ \lim _{x \rightarrow 5} \frac{5-x}{x^{2}-25} $$

Step-by-Step Solution

Answer

The limit as x approaches 5 of the given function is \(-1/10\).

Step 1: Rewrite The Equation

Firstly, rewrite the given equation. It's given as \( \lim _{x \rightarrow 5} \frac{5-x}{x^{2}-25} \)

Step 2: Simplify The Equation

The denominator represents a difference of squares, which can be factored. Hence, \( \lim _{x \rightarrow 5} \frac{5-x}{x^{2}-25} \) can be written as \( \lim _{x \rightarrow 5} \frac{5-x}{(x-5)(x+5)} \)

Step 3: Cancel Out The Similar Terms

Notice the numerators can be rewritten as \(-(x-5)\), which allows us to cancel out the terms \((5-x)\) and \((x-5)\) leaving us with \( \lim _{x \rightarrow 5} \frac{-(x-5)}{(x-5)(x+5)} \) simplifying to \( \lim _{x \rightarrow 5} \frac{-1}{x+5} \)

Step 4: Substitute The Limiting Value

Substitute x=5 into the simplified function on the right side, you get: \( \frac{-1}{5+5} = -\frac{1}{10} \)

Key Concepts

Difference of Squares

The concept of "difference of squares" is crucial in algebra and plays a vital role in calculus, especially when simplifying expressions. It involves expressions in the form \(a^2 - b^2\), which can be factored into \((a-b)(a+b)\). This is because, when expanded, the middle terms cancel each other due to being additive inverses.

For example, consider \(x^2 - 25\). Recognize that it matches the difference of squares format as \(x^2 - 5^2\), and thus we can rewrite it as \((x-5)(x+5)\).

• This factoring makes it easier to simplify expressions, especially when dealing with limits and rational expressions.
• Understanding and applying this concept can make seemingly complex problems much more manageable.

Simplifying Algebraic Expressions

When given an expression, simplifying it usually involves factoring, canceling, and rewriting where necessary to make calculations easier. In calculus, simplifying expressions can turn an indeterminate form into one that is evaluable.

For instance, let's work with the expression \( \frac{5-x}{x^2-25} \). Recognize the denominator as a difference of squares \((x-5)(x+5)\).

• Rewriting the numerator as \(-1(x-5)\) reveals common factors in the numerator and denominator, allowing us to cancel \((x-5)\).
• Always aim to simplify as much as possible before substituting values, which can help avoid undefined expressions or errors.

By simplifying, the expression becomes \( \frac{-1}{x+5} \), making the limit calculation straightforward.

Evaluating Limits by Substitution

In calculus, evaluating limits by substitution can be straightforward after simplification. The goal is to find the value that a function approaches as the variable gets very close to a specific point.

After simplifying a given expression, direct substitution is often the next step. Consider the simplified expression \( \frac{-1}{x+5} \). To find \( \lim_{x \to 5} \frac{-1}{x+5} \), simply replace \(x\) with 5.

• By substituting, we have \( \frac{-1}{5+5} \), which evaluates to \( -\frac{1}{10} \).
• Direct substitution is valid here because the denominator or any division does not result in zero.

Using substitution helps find the actual value of the limit when the function is properly simplified, making the process efficient and reliable.