Problem 20 - A Graphical Approach to Precalculus with Limits

Question

Determine each limit, if it exists. $$\lim _{x \rightarrow 3} \frac{x^{3}-1}{x^{2}+1}$$

Step-by-Step Solution

Answer

The limit is \(\frac{13}{5}\).

Step 1: Substitute the Limit Point

Start by trying to evaluate the limit by directly substituting the value of the approaching point, which is 3, into the expression. Substitute 3 for all occurrences of \(x\): \[\lim _{x ightarrow 3} \frac{x^{3}-1}{x^{2}+1} = \frac{3^3 - 1}{3^2 + 1} = \frac{27 - 1}{9 + 1} = \frac{26}{10}\]

Step 2: Simplify the Expression

Simplify the fraction obtained in Step 1:\[\frac{26}{10} = \frac{13}{5}\]

Step 3: Conclusion

Since the substitution gave us defined values and resulted in a simplified fraction, the limit exists and is \(\frac{13}{5}\).

Key Concepts

Substitution Method

The substitution method involves replacing the variable in a limit expression with a specific numerical value to evaluate the limit of a function. It's a straightforward initial step when evaluating limits, especially when dealing with polynomial functions. For the given exercise, we start by directly substituting the value that the variable approaches, which is 3 in this case.

• Replace each instance of the variable with 3 in the expression \( \lim_{x \rightarrow 3} \frac{x^3 - 1}{x^2 + 1} \).
• Compute the expression: \( \frac{3^3 - 1}{3^2 + 1} = \frac{27 - 1}{9 + 1} = \frac{26}{10} \).

This substitution provides a quick and efficient way to find the limit assuming the function is continuous at that point. If the substitution yields an indeterminate form, like \( \frac{0}{0} \), further analysis along with other methods like factoring, rationalizing, or L'Hôpital's rule might be required.

Simplifying Expressions

Once you've plugged in the values during the substitution step, the resulting expression needs to be simplified to make the solution clearer and more concise. Simplification involves reducing a fraction to its simplest form or reorganizing an equation for easier computation.

• In the problem \( \frac{26}{10} \), the next step is to simplify by finding the greatest common divisor (GCD) of the numerator and the denominator.
• The GCD of 26 and 10 is 2, which allows us to simplify the fraction by dividing both by 2:
• Therefore, \( \frac{26}{10} = \frac{13}{5} \).

This simplification gives a clearer view of the exact result, confirming that the operations yield a definitive answer rather than an approximation. Simplifying expressions not only aids in understanding but is essential for verifying that the limit indeed exists.

Evaluating Limits

Evaluating limits is the process of finding out what value a function approaches as the input approaches a certain point. Limits are foundational in calculus, as they help describe the behavior of functions at points of interest, like near discontinuations or peaks.To successfully evaluate limits:

• Attempt direct substitution to see if it gives a satisfactory answer immediately.
• If the result is not immediately clear, assess the function's form to identify any discontinuities or undefined behaviors.
• In our problem \( \lim_{x \to 3} \frac{x^3 - 1}{x^2 + 1} \), the direct substitution worked well, leading to a computable expression with the result \( \frac{13}{5} \).

This method confirms the limit's existence and reveals the behavior of the function as it approaches the specified variable point. Given the straightforward nature of the function, direct substitution suffices, making evaluation concise and precise. For more complex cases, other limit properties or theorems might become necessary.