Problem 66 - Precalculus

Question

Given \(g(x)=\frac{6 x^{2}-x-1}{2 x-1},\) find $$\lim _{x \rightarrow 1} g(x).$$

Step-by-Step Solution

Answer

The limit is 4.

Step 1: Substitute the Limit Value

First, we substitute \( x = 1 \) directly into the function to check if it's determined or if we need further steps.\[g(1) = \frac{6(1)^2 - 1 - 1}{2(1) - 1} = \frac{6 - 1 - 1}{2 - 1} = \frac{4}{1} = 4.\]

Step 2: Conclusion by Direct Substitution

Since substituting \( x = 1 \) into \( g(x) \) yields a determinate form, we can conclude that the limit is simply the value of \( g(1) \).

Key Concepts

Direct Substitution

The concept of direct substitution is an essential tool in calculus, especially when calculating limits. When you come across a limit problem, the simplest method to use first is direct substitution. This involves plugging the value that the input variable is approaching directly into the function. If the result is a determinate value (like a regular number), then this is the limit of the function.
For example, in the given function \( g(x) = \frac{6x^2 - x - 1}{2x - 1} \) with the limit \( \lim_{x \rightarrow 1} g(x) \), substituting \( x = 1 \) into the function directly results in \( g(1) = 4 \).
Direct substitution is:

• Quick and efficient: Saves time by instantly providing the limit if determinate.
• Possible when the function is continuous at the point of substitution.

When direct substitution yields an indeterminate form, such as \( \frac{0}{0} \), further analysis is needed, often involving algebraic manipulation.

Rational Functions

A rational function is a type of function represented by the quotient of two polynomials. This structure can give rise to unique properties and challenges when working with limits.
Rational functions are expressed as \( \frac{p(x)}{q(x)} \), where \( p(x) \) and \( q(x) \) are polynomials. For the function \( g(x) = \frac{6x^2 - x - 1}{2x - 1} \):

• The numerator is \( 6x^2 - x - 1 \).
• The denominator is \( 2x - 1 \).

These types of functions are continuous everywhere they are defined (i.e., where the denominator is not zero).
Understanding rational functions is crucial because:

• They form many practical models in real-life scenarios.
• They help in developing mathematical intuition around asymptotic behavior and hole locations.

In this exercise, since \( q(1) eq 0 \), the function is continuous and direct substitution works efficiently.

Precalculus Concepts

Mastering limits is part of a series of precalculus concepts crucial to calculus. These are foundational ideas that help set the stage for a deeper understanding of calculus concepts.
Here are some relevant precalculus concepts when studying limits:

• Polynomials and Factoring: Both are vital for simplifying expressions where direct substitution initially fails.
• Continuity: Rational functions are continuous where their denominators aren't zero.

In this exercise, the polynomial in the denominator \( 2x - 1 \) is pivotal. Understanding the condition \( 2x - 1 eq 0 \) assures that for certain x-values (like 1 in our case), \( g(x) \) is continuous.
Strengthening your knowledge of precalculus concepts will aid significantly in identifying when functions are continuous or where more complex limit-solving techniques are necessary.