Problem 54 - Precalculus

Question

Evaluate the limits using the limit properties. $$\lim _{x \rightarrow 3} \frac{\frac{1}{x-2}+3 x^{2}}{x^{2}-6+\frac{1}{x-1}}$$

Step-by-Step Solution

Answer

The limit is 8.

Step 1: Analyze the Expression

First, we identify the expression to evaluate the limit for: \[\lim _{x \rightarrow 3} \frac{\frac{1}{x-2}+3x^{2}}{x^{2}-6+\frac{1}{x-1}}\]We notice there are rational expressions in both the numerator and the denominator that might be undefined for certain values.

Step 2: Substitute Directly into the Limit

Attempt to find the limit by direct substitution of the value 3 into the expression:\[\frac{\frac{1}{3-2}+3 \cdot 3^{2}}{3^{2}-6+\frac{1}{3-1}}\]Simplifying this gives:\[\frac{1 + 27}{3^2 - 6 + \frac{1}{2}}\]which simplifies to:\[\frac{28}{9 + \frac{1}{2} - 6}\].

Step 3: Simplify Denominator

Calculate the denominator:\[9 + \frac{1}{2} - 6 = 3 + \frac{1}{2} = \frac{6}{2} + \frac{1}{2} = \frac{7}{2}\].

Step 4: Simplify the Entire Expression

Now, substitute the simplified denominator back into the expression:\[\frac{28}{\frac{7}{2}}\]which is the same as multiplying by the reciprocal:\[28 \times \frac{2}{7} = 4 \times 2 = 8\].

Step 5: Conclude the Limit Evaluation

Since we successfully evaluated the limit without any indeterminate forms such as \(\frac{0}{0}\), the limit is valid at \(x = 3\).

Key Concepts

Rational Expressions

Rational expressions are essentially fractions where both the numerator and the denominator are polynomials. Think of them as similar to doing math with rational numbers, where numbers are fractions of integers. In the given problem, we have rational expressions both in the numerator and the denominator:

• In the numerator: \( \frac{1}{x-2} + 3x^2 \)
• In the denominator: \( x^2 - 6 + \frac{1}{x-1} \)

Rational expressions can sometimes complicate finding limits, especially if they are undefined at a particular point, such as when the variable reaches a value that makes the denominator zero. In this exercise, we carefully analyze potential points of discontinuity, like \( x = 1 \) and \( x = 2 \), before attempting to substitute \( x = 3 \). This ensures we are not encountering undefined expressions at that specific limit value.

Direct Substitution

Direct substitution is a method of evaluating limits by substituting the variable in the expression with the specific value it's approaching. This method works best when the expression is continuous at the approach value, not leading to undefined conditions.In our problem, we directly substitute \( x = 3 \) into the expression:

• Numerator: \( \frac{1}{3-2} + 3 \times 3^2 = 1 + 27 = 28 \)
• Denominator: \( 3^2 - 6 + \frac{1}{3-1} = 9 - 6 + \frac{1}{2} = 3.5 \)

Through direct substitution, we managed to compute a straightforward limit outcome. This is feasible here as it avoids indeterminate forms, meaning the expression is defined and offers a clear result at \( x = 3 \).

Simplifying Algebraic Expressions

Simplifying algebraic expressions is crucial when working with limits. It involves reducing complex fractions into simpler forms, making it easier to evaluate the expression completely. In our exercise, after direct substitution, we simplify further:

• Simplify the denominator \( 3 + \frac{1}{2} = \frac{6}{2} + \frac{1}{2} = \frac{7}{2} \)
• Then deal with the entire expression: \( \frac{28}{\frac{7}{2}} \) becomes multiplying by the reciprocal: \( 28 \times \frac{2}{7} \)
• This multiplication results in \( 8 \), as the fractions are simplified by reducing common terms.

This approach is crucial for determining limits because a simplified expression often reveals the value of a limit more directly and avoids potential confusion over complex terms. Simplifying often transforms a seemingly difficult limit problem into a simple arithmetic operation.