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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

Expert verified
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.
  • 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 \( \frac{0}{0} \).
  • 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.
  • 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.
Derivatives streamline the limit evaluation process by transforming a complex problem into a more manageable form.

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. By l'Hôpital's Rule, \(\lim _{x \rightarrow 2} \frac{x-2}{x^{2}-1}=\lim _{x \rightarrow 2} \frac{1}{2 x}=\frac{1}{4}\) b. \(\lim _{x \rightarrow 0}(x \sin x)=\lim _{x \rightarrow 0} f(x) g(x)=\lim _{x \rightarrow 0} f^{\prime}(x) \lim _{x \rightarrow 0} g^{\prime}(x)=\) \(\left(\lim _{x \rightarrow 0} 1\right)\left(\lim _{x \rightarrow 0} \cos x\right)=1\) c. \(\lim _{x \rightarrow 0^{+}} x^{1 / x}\) is an indeterminate form. d. The number 1 raised to any fixed power is 1. Therefore, because \((1+x) \rightarrow 1\) as \(x \rightarrow 0,(1+x)^{1 / x} \rightarrow 1\) as \(x \rightarrow 0\) e. The functions \(\ln x^{100}\) and \(\ln x\) have comparable growth rates as \(x \rightarrow \infty\) f. The function \(e^{x}\) grows faster than \(2^{x}\) as \(x \rightarrow \infty\).

Consider the function \(f(x)=\left(a b^{x}+(1-a) c^{x}\right)^{1 / x},\) where \(a, b,\) and \(c\) are positive real numbers with \(0

Locate the critical points of the following functions and use the Second Derivative Test to determine whether they correspond to local maxima, local minima, or neither. $$p(t)=2 t^{3}+3 t^{2}-36 t$$

Find the function \(F\) that satisfies the following differential equations and initial conditions. $$F^{\prime \prime \prime}(x)=672 x^{5}+24 x, F^{\prime \prime}(0)=0, F^{\prime}(0)=2, F(0)=1$$

Given the following acceleration functions of an object moving along a line, find the position function with the given initial velocity and position. $$a(t)=2 e^{-t / 6} ; v(0)=1, s(0)=0$$

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