Problem 43 - College Algebra and Calculus: An Applied Approach

Question

find the limit $$ \lim _{x \rightarrow 2} \frac{x-2}{x^{2}-4 x+4} $$

Step-by-Step Solution

Answer

The limit of the function as \(x \rightarrow 2\) is undefined because the function goes to infinity.

Step 1: Factor The Function

The objective is to eliminate the 0 in the denominator. To do so, factor the function: \(\frac{x-2}{(x-2)(x-2)}\)

Step 2: Cancel The Common Factor

Cancel the common factor from the numerator and the denominator: \( \frac{1}{x-2}\)

Step 3: Direct Substitution

Substitute \(x = 2\) into the simplified function: \( \frac{1}{2-2} \)

Key Concepts

Factoring

Factoring is a fundamental concept in algebra that involves breaking down expressions into their simplest forms. It is especially useful when dealing with polynomials, as it allows us to simplify expressions and solve equations more easily. In the context of limits, factoring can help us eliminate problematic terms, such as those causing a zero in the denominator.

Here, we factor the denominator of the given rational function. The expression \( x^2 - 4x + 4 \) can be rewritten as \((x-2)(x-2)\), because it is a perfect square trinomial. Perfect square trinomials take the form \((a-b)^2 = a^2 - 2ab + b^2\). Recognizing this pattern allows us to simplify our work.

• Identify patterns like \((a-b)^2\) to simplify factoring.
• Factoring helps to cancel out terms creating indeterminate forms like \(\frac{0}{0}\).

After factoring, we have the expression \( \frac{x-2}{(x-2)(x-2)} \). This step is crucial for simplifying the function to proceed with finding the limit.

Rational Functions

Rational functions are expressions formed by the ratio of two polynomials, where the denominator cannot be zero. Understanding rational functions is key in calculus, particularly when exploring limits.

A rational function such as \( \frac{x-2}{x^2 - 4x + 4} \) could become undefined if the denominator equals zero. It is essential to manipulate the expression to avoid division by zero. During our exercise, factoring the denominator reveals it to be \((x-2)(x-2)\), indicating that \(x = 2\) would create a zero denominator, making the limit initially indeterminate as \(\frac{0}{0}\).

• Rational functions can have undefined points where the denominator is zero.
• By factoring, we can often simplify and eliminate these problematic terms.

This simplification allows for further steps, such as cancelling out and applying direct substitution, to find the limit.

Direct Substitution

Direct substitution is a straightforward technique for finding limits. It involves substituting the value into the variable of the function directly. It is the simplest approach but can only be applied if it doesn't create an indeterminate form.

Once we have factored and cancelled common terms in our rational function \( \frac{x-2}{(x-2)(x-2)} \), we are left with \( \frac{1}{x-2} \). Direct substitution of \( x = 2 \) into \( \frac{x-2}{x-2} \) would originally cause \( \frac{0}{0} \), but after cancelling out the \(x-2\) terms, we can safely substitute \(x = 2\) in the expression \( \frac{1}{x-2} \), resulting in \( \frac{1}{2-2} \).

• Direct substitution simplifies finding the limit if no indeterminate forms occur.
• Always simplify the function first, if possible, to avoid \( \frac{0}{0} \) situations.
• Ensure the function remains defined after substitution.

In this case, while substitution shows us that the function approaches infinity, it's important first to ensure that any undefined results have been addressed by simplification.