Problem 5
Question
Explain why \(\lim _{x \rightarrow 3} \frac{x^{2}-7 x+12}{x-3}=\lim _{x \rightarrow 3}(x-4)\) and then evaluate \(\lim _{x \rightarrow 3} \frac{x^{2}-7 x+12}{x-3}\)
Step-by-Step Solution
Verified Answer
Question: Evaluate the limit: \(\lim_{x\rightarrow 3}\frac{x^2-7x+12}{x-3}\).
Answer: -1
1Step 1: Factor the numerator
We need to factor the numerator \(x^2 - 7x + 12\). This can be factored as \((x-3)(x-4)\), as \((x-3)(x-4) = x^2 - 7x + 12\).
2Step 2: Simplify the expression
Now that we have factored the numerator, we can simplify the given expression. We have \(\frac{(x-3)(x-4)}{x-3}\). We can cancel the common terms \((x-3)\) in the numerator and the denominator, resulting in the simpler expression \(x-4\).
3Step 3: Show the equality of limits
Since we have simplified the original expression, we can now show the equality of limits: \(\lim_{x\rightarrow 3} \frac{x^2-7x+12}{x-3} = \lim_{x\rightarrow 3}(x-4)\).
4Step 4: Evaluate the limit
Finally, we can directly evaluate the limit by substituting the value \(x=3\) into the expression \(x-4\). So, \(\lim_{x\rightarrow 3}(x-4) = 3 - 4 = -1\).
Therefore, the evaluated limit is \(\lim _{x \rightarrow 3} \frac{x^{2}-7 x+12}{x-3} = -1\).
Key Concepts
Factoring PolynomialsSimplifying ExpressionsLimit EvaluationCancellation in Rational Expressions
Factoring Polynomials
Factoring polynomials is a critical skill in calculus, vital for simplifying complex algebraic expressions. It involves breaking down a polynomial into a product of simpler factors, which can then be used for further algebraic manipulation or, as in the case of limit problems, to make evaluation possible.
For example, consider the quadratic polynomial \(x^2 - 7x + 12\). To factor this, we look for two numbers that multiply to give the constant term (12) and add up to give the coefficient of the middle term (-7). In this instance, the numbers -3 and -4 fit the criteria, allowing us to rewrite the quadratic as \(x-3)(x-4)\).
Understanding how to factor polynomials is essential when approaching calculus problems because it often reveals cancellation opportunities and can simplify complex expressions. It allows for the assessment of behavior near points that wouldn't be clear from the unfactored form.
For example, consider the quadratic polynomial \(x^2 - 7x + 12\). To factor this, we look for two numbers that multiply to give the constant term (12) and add up to give the coefficient of the middle term (-7). In this instance, the numbers -3 and -4 fit the criteria, allowing us to rewrite the quadratic as \(x-3)(x-4)\).
Understanding how to factor polynomials is essential when approaching calculus problems because it often reveals cancellation opportunities and can simplify complex expressions. It allows for the assessment of behavior near points that wouldn't be clear from the unfactored form.
Simplifying Expressions
Simplifying expressions in calculus not only makes problems more manageable but also clarifies the mathematical landscape of the function or limit in question. After factoring polynomials, simplification often involves the cancellation of like terms.
Let's take the expression \(\frac{(x-3)(x-4)}{x-3}\) as an example. Simplification here involves identifying and eliminating the common factors in the numerator and denominator. In our example, \(x-3\) is present both in the numerator and the denominator, meaning we can cancel it out, leaving us with \(x-4\) as the simpler expression.
Cancellation can only be done with multiplication or division involved—not with addition or subtraction—hence the importance of factoring first. Through simplification, we can avoid unnecessary complications and obtain a clearer path to the solution.
Let's take the expression \(\frac{(x-3)(x-4)}{x-3}\) as an example. Simplification here involves identifying and eliminating the common factors in the numerator and denominator. In our example, \(x-3\) is present both in the numerator and the denominator, meaning we can cancel it out, leaving us with \(x-4\) as the simpler expression.
Cancellation can only be done with multiplication or division involved—not with addition or subtraction—hence the importance of factoring first. Through simplification, we can avoid unnecessary complications and obtain a clearer path to the solution.
Limit Evaluation
Limit evaluation is a cornerstone of calculus, providing insight into the behavior of functions as inputs approach a certain value, often where the function itself might not be clearly defined. The limit of a function as \(x\) approaches a particular value is the value that the function's output gets closer and closer to, as \(x\) gets infinitely close to that value.
In practice, to evaluate \(\lim_{x\rightarrow a}f(x)\), we commonly substitute \(x\) with \(a\) in the function \(f(x)\), provided \(f(x)\) is continuous at \(x=a\). However, when direct substitution leads to an indeterminate form like 0/0, we may need to manipulate the expression, often by factoring and simplifying, to get a form where direct substitution gives a valid result.
In our exercise, once the simplification step has been performed, the process of evaluating the limit becomes straightforward—merely substitute \(x\) with 3 in the simplified expression \(x-4\) to find the limit's value.
In practice, to evaluate \(\lim_{x\rightarrow a}f(x)\), we commonly substitute \(x\) with \(a\) in the function \(f(x)\), provided \(f(x)\) is continuous at \(x=a\). However, when direct substitution leads to an indeterminate form like 0/0, we may need to manipulate the expression, often by factoring and simplifying, to get a form where direct substitution gives a valid result.
In our exercise, once the simplification step has been performed, the process of evaluating the limit becomes straightforward—merely substitute \(x\) with 3 in the simplified expression \(x-4\) to find the limit's value.
Cancellation in Rational Expressions
Cancellation in rational expressions is a method used to simplify expressions that have a polynomial in both the numerator and the denominator. This method relies heavily on the ability to factor polynomials, as seen in our exercise, and aims at reducing the expression to its lowest terms.
In a rational expression like the one we have, \(\frac{x^2-7x+12}{x-3}\), cancellation occurs when the same factor appears in both the numerator and the denominator. Once these common factors have been factored out, they can be divided out of the expression, effectively 'canceling' them, and simplifying the expression to something more manageable. It is vital, however, to note that cancellation can only occur with factors, not terms, a common point of confusion.
Cancellation is especially important when evaluating limits because it often resolves the indeterminate forms that make direct substitution impossible. In our case, it helped transform a potentially complex limit problem into a simple substitution.
In a rational expression like the one we have, \(\frac{x^2-7x+12}{x-3}\), cancellation occurs when the same factor appears in both the numerator and the denominator. Once these common factors have been factored out, they can be divided out of the expression, effectively 'canceling' them, and simplifying the expression to something more manageable. It is vital, however, to note that cancellation can only occur with factors, not terms, a common point of confusion.
Cancellation is especially important when evaluating limits because it often resolves the indeterminate forms that make direct substitution impossible. In our case, it helped transform a potentially complex limit problem into a simple substitution.
Other exercises in this chapter
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