Problem 5
Question
Explain why \(\lim _{x \rightarrow 3} \frac{x^{2}-7 x+12}{x-3}=\lim _{x \rightarrow 3}(x-4)\).
Step-by-Step Solution
Verified Answer
Question: Explain why the statement \(\lim _{x \rightarrow 3} \frac{x^{2}-7 x+12}{x-3} = \lim _{x \rightarrow 3}(x-4)\) is true.
Answer: The statement is true because both expressions have the same limit as \(x\) approaches 3. This is explained by simplifying the first expression by factoring its numerator, which results in an equivalent expression of \(x-4\). When finding the limit of \(x-4\) as \(x\) approaches 3, the result is -1. Since both expressions have the same limit of -1, the statement is true.
1Step 1: Simplify the first expression
Factor the numerator in the first expression:
\(x^{2} - 7x + 12 = (x - 3)(x - 4)\)
Now, substitute the factored form into the original expression:
\(\frac{(x - 3)(x - 4)}{x - 3}\)
Cancel the common term (x-3):
\(x - 4\)
This simplification shows that the given expression can be simplified to \(x-4\).
2Step 2: Find the limits of the given expressions
Now, find the limits of both given expressions as \(x\) approaches 3:
\(\lim _{x \rightarrow 3} \frac{x^{2}-7 x+12}{x-3} = \lim _{x \rightarrow 3}(x-4)\)
When we substitute \(x=3\) into the simplified expression \((x-4)\), we get:
\(3 - 4 = -1\)
Therefore, the limit is:
\(\lim _{x \rightarrow 3}(x-4) = -1\)
3Step 3: Compare the results
Our simplified expression for the first given expression is:
\(\frac{x^{2}-7 x+12}{x-3} = x - 4\)
Since these expressions are equivalent, their limits will also be equivalent:
\(\lim _{x \rightarrow 3} \frac{x^{2}-7 x+12}{x-3} = \lim _{x \rightarrow 3}(x-4) = -1\)
Hence, we have explained why \(\lim _{x \rightarrow 3} \frac{x^{2}-7 x+12}{x-3} = \lim _{x \rightarrow 3}(x-4)\).
Key Concepts
Factorization in AlgebraSimplifying Rational ExpressionsFinding Limits AlgebraicallyLimits of Polynomial Functions
Factorization in Algebra
Factorization is the process of breaking down a complex expression into a product of simpler factors. It's a handy tool in algebra that can simplify expressions and solve equations. For instance, taking a quadratic expression like \(x^{2} - 7x + 12\), we look for two numbers whose product is the constant term, 12, and whose sum is the coefficient of \(x\), which is -7. After finding the pair of numbers, we rewrite the quadratic as \(x^{2} - 7x + 12 = (x - 3)(x - 4)\).
Understanding the factorization process is crucial because it often leads to significant simplifications. When factorization results in a common factor in the numerator and denominator, as seen in rational expressions, it allows for the reduction of the expression to its simplest form. Knowing how to factor is a critical skill for finding limits in calculus, especially when dealing with indeterminate forms that require simplification before applying the limit operation.
Understanding the factorization process is crucial because it often leads to significant simplifications. When factorization results in a common factor in the numerator and denominator, as seen in rational expressions, it allows for the reduction of the expression to its simplest form. Knowing how to factor is a critical skill for finding limits in calculus, especially when dealing with indeterminate forms that require simplification before applying the limit operation.
Simplifying Rational Expressions
Simplifying rational expressions is a fundamental aspect of algebra that involves reducing expressions to their simplest form by eliminating common factors from the numerator and denominator. A rational expression is similar to a fraction, but it contains polynomials in both its numerator and its denominator.
Let's consider the example \(\frac{x^{2}-7x+12}{x-3}\). After factorization, we notice that \(x - 3\) is a common factor in both the numerator and the denominator. We can then simplify the expression by canceling out this common factor, resulting in \(x - 4\). This simplified expression is much easier to handle, particularly when calculating limits, as we've removed the complication of division by zero at \(x=3\).
Let's consider the example \(\frac{x^{2}-7x+12}{x-3}\). After factorization, we notice that \(x - 3\) is a common factor in both the numerator and the denominator. We can then simplify the expression by canceling out this common factor, resulting in \(x - 4\). This simplified expression is much easier to handle, particularly when calculating limits, as we've removed the complication of division by zero at \(x=3\).
Finding Limits Algebraically
Finding limits algebraically involves calculating the value that a function approaches as the input gets close to a certain point. Algebraic manipulation is often utilized to make a function easier to evaluate, especially when a direct substitution results in an undefined expression, such as 0/0.
When we encounter a limit like \(\lim _{x \rightarrow 3} \frac{x^{2}-7x+12}{x-3}\), direct substitution of \(x=3\) is problematic. But algebraic techniques like factorization turn the tide. By factoring the numerator and simplifying the rational expression, we skirt around the direct substitution issue. We then evaluate \(\lim _{x \rightarrow 3}(x-4)\) by simply substituting \(x=3\) into \(x-4\), thereby finding the limit, which is -1.
When we encounter a limit like \(\lim _{x \rightarrow 3} \frac{x^{2}-7x+12}{x-3}\), direct substitution of \(x=3\) is problematic. But algebraic techniques like factorization turn the tide. By factoring the numerator and simplifying the rational expression, we skirt around the direct substitution issue. We then evaluate \(\lim _{x \rightarrow 3}(x-4)\) by simply substituting \(x=3\) into \(x-4\), thereby finding the limit, which is -1.
Limits of Polynomial Functions
When dealing with limits of polynomial functions, we're in luck as they are relatively straightforward to evaluate. Polynomial functions are continuous, which means that their limits at any given point can be found simply by substituting the input value into the function.
In the case of \(\lim _{x \rightarrow 3}(x-4)\), since \(x-4\) is a polynomial function, we directly substitute \(x=3\) to get \(3-4=-1\). This is much simpler than dealing with other types of functions where continuity might be broken or not guaranteed. Hence, the simplicity of polynomial functions underpins much of the work involved in finding limits algebraically and forms a foundational concept within calculus.
In the case of \(\lim _{x \rightarrow 3}(x-4)\), since \(x-4\) is a polynomial function, we directly substitute \(x=3\) to get \(3-4=-1\). This is much simpler than dealing with other types of functions where continuity might be broken or not guaranteed. Hence, the simplicity of polynomial functions underpins much of the work involved in finding limits algebraically and forms a foundational concept within calculus.
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