Problem 13 - Calculus Early Transcendentals

Question

Evaluate the following limits using \(l\) Hópital's Rule. $$\lim _{x \rightarrow 2} \frac{x^{2}-2 x}{8-6 x+x^{2}}$$

Step-by-Step Solution

Answer

Question: Using L'Hôpital's Rule, evaluate the following limit: $$\lim_{x \rightarrow 2} \frac{x^2 - 2x}{8 - 6x + x^2}$$ Answer: -1

Step 1: Check for indeterminate form

To determine if the given limit is in indeterminate form, plug in x = 2 into the expression and see the resulting form: $$\frac{(2)^2 - 2(2)}{8 - 6(2) + (2)^2} = \frac{4 - 4}{8 - 12 + 4} = \frac{0}{0}$$ The result is in the indeterminate form 0/0, so we can proceed to use L'Hôpital's Rule.

Step 2: Differentiate the numerator and the denominator

Using basic differentiation, find the derivatives of the numerator and the denominator: Numerator: \(f(x) = x^2 - 2x\) \(f'(x) = 2x - 2\) Denominator: \(g(x) = 8 - 6x + x^2\) \(g'(x) =- 6 + 2x\) Now the limit can be rewritten as: $$\lim_{x \rightarrow 2} \frac{f'(x)}{g'(x)} = \lim_{x \rightarrow 2} \frac{2x - 2}{-6 + 2x}$$

Step 3: Evaluate the new limit

Using the newly rewritten limit, check again for the indeterminate form by plugging in x = 2: $$\frac{2(2) - 2}{-6 + 2(2)} = \frac{2}{-2}$$ The limit is no longer in indeterminate form. Therefore, the limit is: $$\lim_{x \rightarrow 2} \frac{2x - 2}{-6 + 2x} = \frac{2}{-2} = -1$$ The limit of the given expression using L'Hôpital's Rule is -1.

Key Concepts

Limits

In mathematics, the concept of limits is fundamental when it comes to understanding calculus and continuous functions. The limit of a function describes the behavior of the function as the input approaches a certain value. In this exercise, we need to find the limit of the function \(\lim _{x \rightarrow 2} \frac{x^{2}-2 x}{8-6 x+x^{2}}\). Limits are essential when we deal with points at which a function isn't directly defined.

By evaluating the expression as \(x\) approaches 2, our goal is to assess the trend of the function, rather than the value at that single point. Limits help us determine continuity and differentiability, which are core elements of calculus.

• If the function naturally approaches a particular value, that value is the limit.
• However, if plugging the value leads to division by zero, like in our problem, we might encounter what’s called an indeterminate form.

Indeterminate Forms

The expression \(\frac{0}{0}\) is an example of an indeterminate form. Indeterminate forms occur when an algebraic expression doesn’t straightforwardly lead to a single numerical value. This situation is common when calculating limits. In this exercise, directly plugging \(x = 2\) into the function results in \(\frac{0}{0}\), indicating an indeterminate form.

Indeterminate forms signal that normal evaluation doesn't work and advanced techniques, like L'Hôpital's Rule, need to be applied. L'Hôpital's Rule allows us to differentiate the numerator and the denominator separately until we find a determinate form.

• Typical indeterminate forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), and others.
• Recognizing these forms is crucial for applying calculus techniques correctly.

Differentiation

Differentiation is a process used to find a derivative, which measures how a function changes as its input changes. In this exercise, after identifying the indeterminate form \(\frac{0}{0}\), we use differentiation on both numerator and denominator to apply L'Hôpital's Rule.The initial step requires us to find the derivatives of both the top and bottom parts of the fraction:

• For the numerator, \(f(x) = x^2 - 2x\), the derivative is \(f'(x) = 2x - 2\).
• For the denominator, \(g(x) = 8 - 6x + x^2\), the derivative is \(g'(x) = -6 + 2x\).

Differentiation helps strip down the function to its core components, making it possible to reevaluate the limit:

Using L'Hôpital's Rule, we find:\[ \lim_{x \rightarrow 2} \frac{f'(x)}{g'(x)} = \lim_{x \rightarrow 2} \frac{2x - 2}{-6 + 2x} \]This results in a determinate value of \(-1\) after simplifying, thus resolving the original indeterminate form and finding the limit.