Question

\(\operatorname{Lt}_{x \rightarrow 2} \frac{\sqrt{x^{3}-2 x^{2}+2 x-3}-1}{x-2}=\) (1) 4 (2) 8 (3) 6 (4) 3

Short answer

A. 1 B. 2 C. 3 D. 4 E. The limit does not exist Answer: E. The limit does not exist

Step-by-step solution

Check for indeterminate form

Let's substitute x with 2 into the given function to check for indeterminate form: \(\frac{\sqrt{2^{3}-2\cdot2^{2} + 2\cdot2 - 3}-1}{2-2} = \frac{\sqrt{8-8+4-3}-1}{0} = \frac{0}{0}\) This gives us an indeterminate form. We'll need to manipulate the expression to find the limit.

Apply rationalization

In this situation, we can apply the technique of rationalization to simplify the expression. Multiply the numerator and denominator by the conjugate of the numerator. The conjugate of the numerator is \(\sqrt{x^{3}-2 x^{2} + 2x - 3} + 1\) . \(\lim_{x\rightarrow 2} \frac{\sqrt{x^{3} - 2x^{2} + 2x - 3} - 1}{x - 2} = \lim_{x\rightarrow 2} \frac{ (\sqrt{x^{3} - 2x^{2} + 2x - 3} - 1)\cdot (\sqrt{x^{3} - 2x^{2} + 2x - 3} + 1) }{ (x - 2)\cdot (\sqrt{x^{3} - 2x^{2} + 2x - 3} + 1)}\)

Simplify the expression

Now, we simplify the new expression: \(\lim_{x\rightarrow 2} \frac{ (\sqrt{x^{3} - 2x^{2} + 2x - 3} - 1)\cdot (\sqrt{x^{3} - 2x^{2} + 2x - 3} + 1) }{ (x - 2)\cdot (\sqrt{x^{3} - 2x^{2} + 2x - 3} + 1)} = \lim_{x\rightarrow 2} \frac{ x^3 - 2x^2 + 2x - 3 - 1 + 1 }{ (x - 2)(\sqrt{x^{3} - 2x^{2} + 2x - 3} + 1)}\)

Find the limit

Now that the expression is simplified, let's find the limit as x approaches 2. \(\lim_{x\rightarrow 2} \frac{x^{3} - 2x^{2} + 2x - 3}{(x - 2)(\sqrt{x^{3} - 2x^{2} + 2x - 3} + 1)} = \frac{2^3 - 2(2)^2 + 2(2) - 3}{(2-2)(\sqrt{2^3 - 2(2)^2 + 2(2) - 3} + 1)} = \frac{8 - 8 + 4 - 3}{(0)(1)}\) Since the expression evaluates to \(\frac{1}{0}\), the limit does not exist. Therefore, none of the provided options (1 to 4) are correct.

Key concepts

Indeterminate Forms

When dealing with limits in calculus, we often encounter expressions that are not immediately obvious or solvable. These are known as indeterminate forms. Such forms arise when we evaluate a limit and get a result like \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0\times\infty\), or \(\infty - \infty\), to name a few. These results do not provide any meaningful information as they stand, because they can potentially be simplified to any real number or even become infinite, depending on the context of the function.
Indeterminate forms are especially important because they signal us to dig deeper and apply special techniques to resolve them so that the true limit can be revealed. In our exercise, the substitution of the limit value resulted in \(\frac{0}{0}\), a classic indeterminate form, indicating that further manipulation of the function is necessary to evaluate the limit as x approaches 2.
By recognizing indeterminate forms, we can then proceed with the appropriate mathematical operations – such as rationalization, L'Hôpital's rule, or algebraic manipulation – to discover the behavior of the function near the point of interest. In this case, the indeterminate form serves as a clue that prompted us to rationalize the expression.

Rationalization Technique

When faced with a limit that results in an indeterminate form, like \(\frac{0}{0}\), we can use the rationalization technique to make the function more manageable. Rationalization involves multiplying the numerator and denominator by a “conjugate” to eliminate radicals or other troublesome components.
For our example, we rationalized the numerator by multiplying it with its conjugate, which is essentially the same expression with the opposite middle sign (in this case, from minus to plus). This is a valuable step in handling radical expressions where direct substitution leads to indeterminate forms.

Why Rationalize?

• Rationalization can reveal simplified forms that allow us to cancel out terms.
• It helps in removing roots and complex terms in the numerator or denominator.
• It provides a path forward where direct substitution fails.

In summary, rationalization is a useful algebraic tool that aids in the computation of limits by transforming the given function into a form that can be simplified and hence, resolved, as shown in steps 2 and 3 of our solution.

Limits in Calculus

The concept of limits is foundational in calculus. A limit describes the behavior of a function as the input approaches a certain value, which can be a number or infinity. Limits help in understanding the behavior of functions especially when they aren't well-defined at that point due to discontinuity, a hole, or a vertical asymptote.
The notion of a limit is what allows us to define derivatives and integrals, which are core to the study of calculus. Calculating a limit may involve straightforward substitution if the function is continuous at the point in question. However, if the direct substitution results in an indeterminate form, additional techniques are necessary.
In the context of our exercise, we sought the limit as x approaches 2. Direct substitution gave an indeterminate form, so we employed rationalization as a technique to process the limit. However, after simplifying, we found that the limit does not exist as the resulting expression approached the form \(\frac{1}{0}\), which is undefined in mathematics. This demonstrates the importance of evaluating limits properly and understanding that not all expressions will conform to expected outcomes.