Problem 30 - Calculus and its applications

Question

The initial substitution of \(x=a\) yields the form \(0 / 0 .\) Look for ways to simplify the function algebraically, or use a table or graph to determine the limit. When necessary, state that the limit does not exist. $$ \lim _{x \rightarrow-1} \frac{x^{2}+5 x+4}{x^{2}+2 x+1} $$

Step-by-Step Solution

Answer

The limit does not exist (\(\frac{3}{0}\)).

Step 1: Identify the Function Structure

To simplify the limit \( \lim _{x \rightarrow -1} \frac{x^2 + 5x + 4}{x^2 + 2x + 1} \), we first recognize that when substituting \(x = -1\), both the numerator \(x^2 + 5x + 4\) and denominator \(x^2 + 2x + 1\) become zero, resulting in an indeterminate form \(\frac{0}{0}\). Our goal now is to simplify the expression.

Step 2: Factor the Numerator

We factor the numerator \(x^2 + 5x + 4\). We need two numbers that multiply to 4 and add to 5. These numbers are 4 and 1. Thus, the numerator can be factored as \((x+4)(x+1)\).

Step 3: Factor the Denominator

Next, we factor the denominator \(x^2 + 2x + 1\). We need two numbers that multiply to 1 and add to 2. These numbers are 1 and 1. Thus, the denominator can be factored as \((x+1)(x+1)\) or \((x+1)^2\).

Step 4: Simplify the Expression

The expression becomes \(\frac{(x+4)(x+1)}{(x+1)(x+1)}\). We cancel the common factor \((x+1)\) from both the numerator and the denominator, resulting in the simplified expression \(\frac{x+4}{x+1}\).

Step 5: Evaluate the Limit

Substitute \(x = -1\) into the simplified expression \(\frac{x+4}{x+1}\), which gives us \(\frac{-1+4}{-1+1}\). This results in \(\frac{3}{0}\), indicating the limit does not exist, since division by zero is undefined.

Key Concepts

Indeterminate Forms

When you first evaluate a limit by substituting a value and end up with \(\frac{0}{0}\), you have encountered an indeterminate form. Indeterminate forms arise because limits might tend toward more than one value, due to a balancing effect in the numerator and denominator.
This essentially means direct substitution doesn't provide a clear answer. Fortunately, there are strategies to resolve this. One common method is to simplify the expression using algebraic techniques like factoring. Another approach involves using graphs or tables to understand behavior near the limit point.
Mastering indeterminate forms is crucial for calculus, as they often appear in problems requiring deeper analysis beyond straightforward evaluation. Remember, each indeterminate form is a signal to explore the expression further rather than being a dead-end.

Factoring Polynomials

Polynomials often turn chaotic expressions into simpler forms through factoring. Let's say you have a polynomial like \(x^2 + 5x + 4\).

• Find pairs of numbers that multiply to the constant term (here, 4).
• Ensure these numbers add up to the coefficient of the linear term (here, 5).

For \(x^2 + 5x + 4\), numbers 4 and 1 work perfectly, giving you \( (x+4)(x+1) \).Similarly, factoring the denominator \(x^2 + 2x + 1\) is about simplifying it to its roots. Here, the numbers 1 and 1 multiply to 1 and sum to 2, so it factors as \( (x+1)(x+1) \) or \( (x+1)^2 \).Knowing how to factor polynomials is critical because it unlocks expressions that appear as tricky hurdles, transforming them into more workable equations.

Simplifying Rational Expressions

Simplifying rational expressions is a valuable skill that helps in examining functions closely. It's like peering through clutter to see the core components clearly. Often, the process involves:

• Factoring the numerator and the denominator of the expression.
• Cancelling out any common factors that appear in both parts.

In our example, after factoring, the expression simplifies from \( \frac{(x+4)(x+1)}{(x+1)(x+1)} \) to \( \frac{x+4}{x+1} \) by cancelling the common \( (x+1) \).This simplification can change an indeterminate form into a more pronounced one, or give clearer insight into the behavior near the limit. Always keep in mind, cancelling terms require that the terms are non-zero. If you're simplifying a limit, this ensures you're approaching a proper evaluation point without introducing division by zero.