Problem 14
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. $$\lim _{x \rightarrow 2} \frac{x^{2}-5 x+6}{x-2}$$
Step-by-Step Solution
Verified Answer
The limit is -1.
1Step 1: Factor the Numerator
The expression in the limit is \( \frac{x^2 - 5x + 6}{x - 2} \). Start by factoring the quadratic expression in the numerator: \( x^2 - 5x + 6 \). This can be factored into \( (x - 2)(x - 3) \).
2Step 2: Simplify the Expression
Substitute the factorization back into the original expression: \( \frac{(x - 2)(x - 3)}{x - 2} \). Now, cancel out the common factor \( (x - 2) \) in the numerator and denominator, which simplifies the expression to \( x - 3 \).
3Step 3: Evaluate the Limit
Now that the expression is simplified to \( x - 3 \), substitute \( x = 2 \) into the expression: \( 2 - 3 = -1 \). Therefore, \( \lim_{x \rightarrow 2} \frac{x^2 - 5x + 6}{x - 2} = -1 \).
Key Concepts
Algebraic ManipulationFactoring QuadraticsRational Expressions
Algebraic Manipulation
Algebraic manipulation is a crucial skill in calculus, especially when working with limits. It involves rearranging expressions to make them easier to work with or solve. This might involve factoring, expanding, adding, subtracting, or cancelling terms. In limits, algebraic manipulation often helps to identify removable discontinuities, which are points that make the original form of an expression undefined but become defined upon simplification.
Let's consider the problem at hand. The original expression is \( \frac{x^2 - 5x + 6}{x - 2} \), which is undefined at \( x = 2 \) because the denominator becomes zero. By factoring the numerator, we can simplify the expression and "remove" this discontinuity.
These kinds of manipulations are not just tricks but are fundamental techniques that allow mathematicians to handle the intricate calculations in calculus.
Let's consider the problem at hand. The original expression is \( \frac{x^2 - 5x + 6}{x - 2} \), which is undefined at \( x = 2 \) because the denominator becomes zero. By factoring the numerator, we can simplify the expression and "remove" this discontinuity.
These kinds of manipulations are not just tricks but are fundamental techniques that allow mathematicians to handle the intricate calculations in calculus.
Factoring Quadratics
Factoring quadratics is a method of breaking down a quadratic expression into the product of its linear factors. It is a key technique, particularly when dealing with limits and algebraic simplification. For a quadratic expression such as \( x^2 - 5x + 6 \), our goal is to express it as \( (x - a)(x - b) \), where \( a \) and \( b \) are roots of the equation.
Factoring is performed by finding two numbers whose product is equal to the constant term (here, 6) and whose sum equals the linear coefficient (here, -5). The numbers -2 and -3 fit these criteria because \(-2 \times -3 = 6\) and \(-2 + (-3) = -5\). Thus, \( x^2 - 5x + 6 \) factors to \((x - 2)(x - 3)\).
This factorization reveals the roots of the quadratic and, in the context of limits, helps cancel out terms that cause undefined expressions, making it possible to proceed with limit evaluation.
Factoring is performed by finding two numbers whose product is equal to the constant term (here, 6) and whose sum equals the linear coefficient (here, -5). The numbers -2 and -3 fit these criteria because \(-2 \times -3 = 6\) and \(-2 + (-3) = -5\). Thus, \( x^2 - 5x + 6 \) factors to \((x - 2)(x - 3)\).
This factorization reveals the roots of the quadratic and, in the context of limits, helps cancel out terms that cause undefined expressions, making it possible to proceed with limit evaluation.
Rational Expressions
Rational expressions are fractions in which both the numerator and the denominator are polynomials. Handling these expressions involves both understanding how they can be simplified and maintaining their integrity as fractions.
In the context of limits, simplifying rational expressions often reveals where a limit exists. When we have an expression like \( \frac{(x - 2)(x - 3)}{x - 2} \), the factor \( x - 2 \) causes the expression to be undefined at \( x = 2 \).
However, by cancelling the \( x - 2 \) terms, the function simplifies to \( x - 3 \), which is defined for all real numbers, including \( x = 2 \). Thus, the limit can be found by direct substitution into the simplified expression, leading to the conclusion \( \lim_{x \rightarrow 2} = -1 \). This process highlights the importance of simplifying rational expressions to solve limit problems, as it helps in removing the discontinuities.
In the context of limits, simplifying rational expressions often reveals where a limit exists. When we have an expression like \( \frac{(x - 2)(x - 3)}{x - 2} \), the factor \( x - 2 \) causes the expression to be undefined at \( x = 2 \).
However, by cancelling the \( x - 2 \) terms, the function simplifies to \( x - 3 \), which is defined for all real numbers, including \( x = 2 \). Thus, the limit can be found by direct substitution into the simplified expression, leading to the conclusion \( \lim_{x \rightarrow 2} = -1 \). This process highlights the importance of simplifying rational expressions to solve limit problems, as it helps in removing the discontinuities.
Other exercises in this chapter
Problem 14
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