Problem 63 - Calculus Early Transcendentals

Question

Evaluate the following limits, where \(c\) and \(k\) are constants. \(\lim _{x \rightarrow 1} \frac{\sqrt{10 x-9}-1}{x-1}\)

Step-by-Step Solution

Answer

Answer: The limit of the function as \(x\) approaches \(1\) is \(0\).

Step 1: Apply rationalization by multiplying the numerator and denominator by the conjugate of the numerator

First, let's multiply both the numerator and the denominator by the conjugate of the numerator, \((\sqrt{10x-9} + 1)\): \(\lim _{x \rightarrow 1} \frac{\sqrt{10 x-9}-1}{x-1} \cdot \frac{\sqrt{10x-9}+1}{\sqrt{10x-9}+1}\)

Step 2: Simplify the expression

Now let's multiply and simplify the expression: \(\lim _{x \rightarrow 1} \frac{(10x - 9) - 2\sqrt{10x - 9} + 1}{(x-1)(\sqrt{10x-9}+1)}\) \(\lim _{x \rightarrow 1} \frac{10x - 8 - 2\sqrt{10x - 9}}{(x-1)(\sqrt{10x-9}+1)}\)

Step 3: Factor out a constant from the numerator

Next, let's factor out \(2\) from the numerator: \(\lim _{x \rightarrow 1} \frac{2(5x-4-\sqrt{10x-9})}{(x-1)(\sqrt{10x-9}+1)}\)

Step 4: Apply the limit and simplify

Now we'll apply the limit \(x \rightarrow 1\) to the expression. We know that when \(x=1\), the denominator will no longer be \(0\): \(\lim _{x \rightarrow 1} \frac{2(5x-4-\sqrt{10x-9})}{(x-1)(\sqrt{10x-9}+1)} = \frac{2(5(1)-4-\sqrt{10(1)-9})}{((1)-1)(\sqrt{10(1)-9}+1)}\) Simplify the expression: \(\frac{2(1-\sqrt{1})}{(0)(\sqrt{1}+1)} = \frac{2(0)}{0} = \boxed{0}\) The limit of the function as \(x\) approaches \(1\) is \(0\).

Key Concepts

Rationalization

Rationalization is a useful technique in calculus, particularly when dealing with limits involving square roots. The goal is to eliminate the square root from the numerator or denominator of a fraction by multiplying by a suitable expression. This helps in simplifying the evaluation of limits.

An expression like \(\frac{\sqrt{10x-9}-1}{x-1}\) becomes simpler when you multiply both the numerator and the denominator by the conjugate of the numerator. The conjugate in this case is \(\sqrt{10x-9}+1\).

• This removes the square root from the numerator.
• It helps reveal simpler expressions by using the difference of squares.

Rationalization is a powerful method to turn complex expressions into manageable ones, making limit evaluation more straightforward.

Conjugate Multiplication

Conjugate multiplication plays a crucial role when dealing with expressions containing square roots. By using the conjugate, we can simplify expressions and remove radicals.

In our limit problem, the original expression \(\sqrt{10x-9} - 1\) was paired with its conjugate, \(\sqrt{10x-9} + 1\), for multiplication. This has the effect of creating a difference of squares:

• This transforms the expression into \((10x-9) - 1^2\), which simplifies to \(10x-10\).
• By using conjugate multiplication, the square root disappears and the polynomial form takes over, which is easier to handle when finding limits.

Conjugates are not just about removing roots, but they also unveil opportunities to factor further down the line.

Limit Simplification

Simplifying limits is about making expressions as straightforward as possible so that substitution can be used. After rationalization and conjugate multiplication, the limit expression needs to be further simplified.

In our problem, we ended up with \(\lim _{x \rightarrow 1} \frac{10x-8-2\sqrt{10x-9}}{(x-1)(\sqrt{10x-9}+1)}\). By factoring and simplifying terms:

• We factored out a 2 from the numerator, yielding an easier expression to work with.
• This sets up the next step of directly applying the limit \(x \rightarrow 1\), simplifying further until the numerator itself resolves to zero.

This iterative process of simplification is key in handling complex expressions and finding limits accurately.

Numerator and Denominator Manipulation

Manipulating the numerator and denominator to solve limits can start off complex but becomes insightful with practice.

In the problem, after multiplying by the conjugate, both parts of the fraction needed refinement:

• We transformed \(10x - 8 - 2\sqrt{10x - 9}\) in the numerator through logical factoring.
• This allowed the elimination of indeterminate forms, leading to the direct calculation of limits as \(x\) approached 1.

The bottom line is to adjust each part of the fraction to clear any ambiguities or undefined expressions, allowing for a smooth limit evaluation. Recognizing patterns and strategic manipulation are essential skills here.