Problem 75 - University Calculus: Early Transcendentals

Question

Which one is correct, and which one is wrong? Give reasons for your answers. $$\text { a. } \lim _{x \rightarrow 3} \frac{x-3}{x^{2}-3}=\lim _{x \rightarrow 3} \frac{1}{2 x}=\frac{1}{6}$$.$$\text { b. } \lim _{x \rightarrow 3} \frac{x-3}{x^{2}-3}=\frac{0}{6}=0$$.

Step-by-Step Solution

Answer

a is correct; b is wrong.

Step 1: Evaluate the Limit Expression

Let's evaluate \( \lim _{x \rightarrow 3} \frac{x-3}{x^{2}-3} \). This expression simplifies to \( \frac{x-3}{(x-\sqrt{3})(x+\sqrt{3})} \) as \( x \rightarrow 3 \), both the numerator and one factor in the denominator, \( x-\sqrt{3} \), approach zero. This indicates an indeterminate form \( \frac{0}{0} \), so we must do further algebraic manipulation to resolve the limit.

Step 2: Simplify the Algebraic Expression

Factor \( x^2 - 3 \) as \( (x-\sqrt{3})(x+\sqrt{3}) \). Therefore, \( \lim _{x \rightarrow 3} \frac{x-3}{x^{2}-3} \) becomes \( \lim _{x \rightarrow 3} \frac{x-3}{(x-3)(x+3)} \). Simplifying this, we get \( \lim _{x \rightarrow 3} \frac{1}{x+3} \).

Step 3: Calculate the Limit

Now, substitute \( x = 3 \) into the expression \( \frac{1}{x+3} \). Therefore, the limit becomes \( \frac{1}{3+3} = \frac{1}{6} \).

Step 4: Verify Each Option

a. According to the calculation, \( \lim _{x \rightarrow 3} \frac{x-3}{x^{2}-3} = \frac{1}{6} \) is correct. The method of substitution directly after factorization confirms its correctness. b. The expression \( \frac{x-3}{x^2-3} = \frac{0}{6} \) is incorrect as it omits the step of resolving the indeterminate form. Direct substitution without simplifying gives a misleading result.

Key Concepts

Indeterminate Forms

When dealing with limits, we often encounter expressions that don't immediately simplify to a recognizable number, such as \( \frac{0}{0} \). This is known as an indeterminate form, which occurs when both the numerator and denominator of a fraction approach zero as \( x \) approaches a particular value. In these cases, simply substituting the value into the expression won't give us the answer we're looking for.

• Since the limit expression becomes \( \frac{0}{0} \), further analysis is needed to determine the limit's actual value.
• Identifying indeterminate forms is crucial because they signal that algebraic manipulation or other methods must be used to find the correct limit.
• Indeterminate forms include other combinations like \( \frac{\infty}{\infty} \) and forms like \( 0 \cdot \infty \), \( \infty - \infty \).

Understanding when you're dealing with an indeterminate form is the first step in choosing the right approach to solve the limit problem.

Algebraic Manipulation

Algebraic manipulation is critical when resolving indeterminate forms in limits. This process involves rewriting the algebraic expression in a way that resolves the indeterminacy. We see this in action by breaking down the original limit expression:\[ \lim_{x \rightarrow 3} \frac{x-3}{x^2-3} \].

• First, factor the denominator: \( x^2 - 3 \) can be expressed as \( (x-\sqrt{3})(x+\sqrt{3}) \) or \( (x-3)(x+3) \) depending on the relevant values.
• Then, cancel out the common term \((x-3)\) in the numerator and the denominator.
• This simplification changes the expression to a form that is not indeterminate, allowing you to find the limit with direct substitution.

By applying algebraic manipulation correctly, you transform the expression to one that can be easily managed, ultimately revealing the correct limit value.

Factorization

Factorization is a fundamental technique in simplifying algebraic expressions, especially in resolving indeterminate forms. It involves expressing a polynomial or expression as a product of its factors, which simplifies the process of determining limits.

• In the given problem, factorizing \( x^2 - 3 \) was essential. By recognizing it as \( (x-\sqrt{3})(x+\sqrt{3}) \) or equivalently \( (x-3)(x+3) \), we could adequately rework the expression for the limit.
• Factorization allows you to cancel common terms in the numerator and the denominator, which helps remove the indeterminate portion of the expression.
• This process can often transform an otherwise complex problem into a simpler one, making calculations straightforward.

Mastering factorization can greatly ease the process of evaluating limits, especially when faced with complex polynomial expressions.