Chapter 4: Problem 13
Evaluate the following limits using l' Hôpital's Rule. $$\lim _{x \rightarrow 2} \frac{x^{2}-2 x}{8-6 x+x^{2}}$$
Short Answer
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Question: Evaluate the limit of the function as x approaches 2: $$\lim _{x \rightarrow 2} \frac{x^{2}-2 x}{8-6 x+x^{2}}$$
Answer: The limit of the given function as x approaches 2 is -1.
Step by step solution
01
Direct Substitution
Let's try to directly substitute the value x =2 into the given function:
$$\frac{(2)^{2}-2(2)}{8-6(2)+(2)^{2}} = \frac{4 - 4}{8 -12 +4}$$
Here, we get 0 in both numerator and denominator, which indicates an indeterminate form 0/0. Therefore, we need to apply L'Hôpital's Rule to find the limit.
02
Applying L'Hôpital's Rule
First, we find the derivatives of the numerator and the denominator with respect to x:
Numerator:
$$\frac{d}{dx}(x^2 - 2x) = 2x - 2$$
Denominator:
$$\frac{d}{dx}(8 - 6x + x^2) = -6 + 2x$$
Now, we find the limit of the ratio of these derivatives as x approaches 2:
$$\lim _{x \rightarrow 2} \frac{2x - 2}{-6 + 2x}$$
03
Direct Substitution of Derivatives
Now, substitute the value x = 2 into the derivatives:
$$\frac{2(2) - 2}{-6 + 2(2)} = \frac{4 - 2}{-6 + 4}$$
The expression simplifies to:
$$\frac{2}{-2} = -1$$
Since we're able to find the limit without any indeterminate form, we've successfully found the limit using L'Hôpital's Rule.
04
Final Answer
The limit of the given function as x approaches 2 using L'Hôpital's Rule is:
$$\lim _{x \rightarrow 2} \frac{x^{2}-2 x}{8-6 x+x^{2}} = -1$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limits
Limits are a fundamental concept in calculus that describe the behavior of a function as its input approaches a particular value. They help us understand what a function is approaching, even if it never actually reaches that point. Suppose we have a function \( f(x) \) and we are interested in knowing what happens to \( f(x) \) as \( x \) approaches a certain number, say \( a \). We write this as \( \lim_{x \to a} f(x) \).
In some cases, limits are straightforward to evaluate by direct substitution. However, if direct substitution leads to an expression like \( \frac{0}{0} \), then we encounter what's called an 'indeterminate form', which suggests the need for a different approach.
In some cases, limits are straightforward to evaluate by direct substitution. However, if direct substitution leads to an expression like \( \frac{0}{0} \), then we encounter what's called an 'indeterminate form', which suggests the need for a different approach.
- When \( x \rightarrow 2 \) in \( \lim_{x \rightarrow 2} \frac{x^{2}-2 x}{8-6 x+x^{2}} \), direct substitution results in \( \frac{0}{0} \), making it clear that simple substitution won't work, indicating an indeterminate form.
- This situation requires a more advanced technique such as L'Hôpital's Rule to evaluate the limit properly.
Indeterminate Forms
Indeterminate forms arise in calculus when attempting to evaluate a limit leads to an expression that's not immediately solvable. Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), among others. These forms signal that simple arithmetic operations aren't enough to find the limit value.
For example, evaluating \( \lim _{x \rightarrow 2} \frac{x^{2}-2 x}{8-6 x+x^{2}} \) directly results in the form \( \frac{0}{0} \).
For example, evaluating \( \lim _{x \rightarrow 2} \frac{x^{2}-2 x}{8-6 x+x^{2}} \) directly results in the form \( \frac{0}{0} \).
- This specific form \( \frac{0}{0} \) tells us that both the numerator and the denominator become zero. As a result, we're unable to determine the limit using basic algebraic methods.
- To further analyze such expressions, we turn to methods like L'Hôpital's Rule or algebraic manipulation to find derivatives, which will potentially simplify the limit problem.
Derivatives
Derivatives provide a key tool in calculus for analysing the behavior of functions. They measure the rate at which a function changes as its input changes. Mathematically, the derivative of a function \( f \) with respect to \( x \) is denoted \( f'(x) \) or \( \frac{df}{dx} \).
In the context of evaluating limits, derivatives are essential when applying L'Hôpital's Rule. This rule states that if the limit \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form is encountered, and if the derivatives \( f'(x) \) and \( g'(x) \) exist, then \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \), provided the limit on the right-hand side exists.
In the context of evaluating limits, derivatives are essential when applying L'Hôpital's Rule. This rule states that if the limit \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form is encountered, and if the derivatives \( f'(x) \) and \( g'(x) \) exist, then \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \), provided the limit on the right-hand side exists.
- For the problem \( \lim _{x \rightarrow 2} \frac{x^{2}-2 x}{8-6 x+x^{2}} \), we use derivatives: the numerator’s derivative is \( 2x - 2 \) and the denominator’s is \( -6 + 2x \).
- Substituting \( x = 2 \) to the derivatives \( \lim _{x \rightarrow 2} \frac{2x - 2}{-6 + 2x} \) helps us find the limit as \( -1 \), showing that the evaluation is much simpler post-derivation.