Problem 31 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

Option b is correct, option a is wrong.

Step 1: Analyze Expression Limits

Consider the limit \( \lim_{x \rightarrow 3} \frac{x-3}{x^2-3} \). We will examine whether substituting 3 into the expression gives a determinate value or results in an indeterminate form.

Step 2: Substitution Check

Substitute \( x = 3 \) into the expression: \( \frac{3-3}{3^2-3} = \frac{0}{6} \). Direct substitution shows this is not an indeterminate form like \( \frac{0}{0} \).

Step 3: Evaluate Limit a

In option a, rewrite the expression as \( \lim_{x \rightarrow 3} \frac{1}{2x} \). If \( x = 3 \), then \( \frac{1}{2 \times 3} = \frac{1}{6} \). Therefore, the simplification here does not match the original limit.

Step 4: Evaluate Limit b

According to option b, \( \lim_{x \rightarrow 3} \frac{x-3}{x^2-3} = 0 \). Since \( \frac{3-3}{9-3} = \frac{0}{6} = 0 \), option b is indeed correct.

Step 5: Conclusion

Option a recalculates the limit incorrectly as it assumes \( \lim_{x \rightarrow 3} \frac{1}{2x} = \frac{1}{6} \), which is not the result of the original limit expression \( \frac{x-3}{x^2-3} \). Option b correctly evaluates the expression as 0.

Key Concepts

Indeterminate Forms

When dealing with limits in calculus, it's crucial to understand the concept of indeterminate forms. Indeterminate forms occur when substituting a value into a mathematical expression results in an ambiguous expression like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms do not signify a specific number, meaning you cannot directly evaluate them by simple substitution.
For example, if you try plugging in \(x = 3\) directly into \( \frac{x-3}{x^2-3} \), you obtain \( \frac{0}{6} \), which is not indeterminate. This implies that a numerical value can be calculated directly, making it critical to always assess an expression for indeterminacy first before performing further calculations.
The common indeterminate forms to watch out for include:

• \( \frac{0}{0} \)
• \( \infty - \infty \)
• \( 0 \cdot \infty \)
• \( \frac{\infty}{\infty} \)

These forms require additional techniques like factoring, conjugates, or L'Hopital's Rule to resolve.

Limit Evaluation

Limit evaluation is a fundamental technique in calculus used to determine the behavior of a function as it approaches a particular point. The process involves analyzing the expression and deciding the best method to evaluate the limit, such as substitution, factoring, or other algebraic methods.
In our problem, we determined that direct substitution of \( x = 3 \) into \( \frac{x-3}{x^2-3} \) yielded a determinate value because it simplified to \( \frac{0}{6} \). This means the limit can be directly evaluated without any need for further simplification. Confirming whether a limit results in a determinate or indeterminate form can guide you on how to proceed appropriately.
Always remember, when evaluating limits:

• Check if direct substitution solves the limit.
• If a determinate form occurs, simplify further if necessary.
• Use algebraic techniques if you face indeterminate forms.

By applying these steps smartly, you'll gain confidence in handling broad types of limit problems.

Substitution Method

The substitution method in calculus is often the first technique used in limit evaluation. It involves directly plugging the value of \( x \) into a function to see if the limit can be determined outright from a simple calculation.
This method is straightforward when substituting results in a clear, finite number as in our example, \( \frac{x-3}{x^2-3} \), where substituting \( x = 3 \) led directly to \( \frac{0}{6} \), confirming the validity of option b with a limit evaluation outcome of 0.
In the substitution method:

• Insert the limiting value into the original function.
• Check the form of the expression to see if it is determinate.
• If indeterminate, try simplifying the expression first or apply advanced techniques.

Always begin with substitution—if it works, great, if not, you will know to try other methods.