Question

find the indicated limit or state that it does not exist. In many cases, you will want to do some algebra before trying to evaluate the limit. $$ \lim _{x \rightarrow 2} \frac{x^{2}-5 x+6}{x-2} $$

Short answer

The limit is -1.

Step-by-step solution

Identify the Function to Analyze

The function given is \( \frac{x^2 - 5x + 6}{x - 2} \). We need to find the limit as \( x \) approaches 2.

Substitute and Check for Indeterminate Form

Substitute \( x = 2 \) directly into the function: \( \frac{2^2 - 5(2) + 6}{2 - 2} = \frac{4 - 10 + 6}{0} \), which simplifies to \( \frac{0}{0} \), an indeterminate form. This indicates that direct substitution is not sufficient, and algebraic manipulation is necessary.

Factor the Numerator

Factor the quadratic \( x^2 - 5x + 6 \), which can be expressed as \((x - 2)(x - 3)\).

Simplify the Expression

Rewrite the function using the factored form: \( \frac{(x-2)(x-3)}{x-2} \). Cancel out the common factor \( (x-2) \) from the numerator and denominator, simplifying the expression to \( x - 3 \).

Evaluate the Limit of the Simplified Expression

With the simplified function \( x - 3 \), substitute \( x = 2 \) directly into the expression: \( 2 - 3 \), which equals \( -1 \).

Conclude the Solution

The limit of the original function \( \lim_{x \to 2} \frac{x^2 - 5x + 6}{x - 2} \) is \( -1 \).

Key concepts

Indeterminate Forms

When you're dealing with limits, you'll often encounter something called an "indeterminate form." This arises when a substitution leads to a result that is undefined or indeterminate, such as \( \frac{0}{0} \). Such forms signal that direct substitution in a limit problem doesn't work directly, and more algebra is needed. In our example with \( \lim _{x \rightarrow 2} \frac{x^{2}-5 x+6}{x-2} \), substituting \( x = 2 \) directly gives \( \frac{0}{0}\), an indeterminate form.

To resolve this, algebraic techniques like factoring, rationalizing the numerator, or other methods are employed to simplify the expression so that a meaningful limit can be evaluated. Basically, an indeterminate form is a red flag that more work is necessary to find a limit.

Factoring Quadratics

Factoring quadratics is an essential algebraic skill, particularly when solving limit problems. A quadratic is a polynomial of degree two, typically in the form \( ax^2 + bx + c \). When finding limits involving quadratics, recognizing that the expression can be factored is key.

In the example \( x^2 - 5x + 6 \), we see it can be factored into \((x-2)(x-3)\).

• Look for factors of the constant term (here 6) that add up to the coefficient of the linear term (here -5).
• In this case, -2 and -3 work perfectly.

Factoring not only simplifies the expression but can eliminate indeterminate forms by canceling common factors, paving the way for easier limit evaluation.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. Such functions often have a lot going on in terms of limits. With rational functions, limits can sometimes result in indeterminate forms, as we've seen.

Understanding the behavior of both the numerator and denominator as \( x \) approaches a certain value is crucial. For \( \frac{x^2 - 5x + 6}{x-2} \), both the numerator and denominator become zero as \( x \) approaches 2, which is problematic.

This means factoring (if possible) to cancel terms or use other algebraic methods to simplify the expression becomes necessary for finding the limit without indeterminacy. Rational functions often require one to discern patterns, manipulate the expression creatively, and possibly find holes, asymptotes, or vertical tangents.

Simplifying Expressions

Simplifying expressions is a vital step in resolving limits, especially when confronting indeterminate forms. The main aim is to rewrite the given expression in a simpler, yet equivalent form that avoids the problem form.

In our specific case of \( \frac{x^2 - 5x + 6}{x-2} \), we factor the numerator into \((x - 2)(x - 3)\). This allows us to cancel the \((x-2)\) term from both the numerator and denominator, simplifying the expression to \(x - 3\).

After simplification, checking by substitution becomes feasible without landing back at an indeterminate form. Simplifying the expression effectively transforms a complicated problem into a simple one, enabling straightforward computation of the desired limit. By doing this, we arrived at the limit \(-1\) easily and accurately. By mastering simplification techniques, algebraic manipulation becomes a useful tool in solving many types of mathematical problems.