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Chapter 7: Problem 76

Which one is correct, and which one is wrong? Give reasons for your answers. $$\begin{aligned} a. \lim _{x \rightarrow 0} \frac{x^{2}-2 x}{x^{2}-\sin x} &=\lim _{x \rightarrow 0} \frac{2 x-2}{2 x-\cos x} \\ &=\lim _{x \rightarrow 0} \frac{2}{2+\sin x}=\frac{2}{2+0}=1 \end{aligned}$$ $$\text { b. } \lim _{x \rightarrow 0} \frac{x^{2}-2 x}{x^{2}-\sin x}=\lim _{x \rightarrow 0} \frac{2 x-2}{2 x-\cos x}=\frac{-2}{0-1}=2$$

Short Answer

Expert verified
Part a is wrong; it incorrectly calculates the final limit. Part b is correct, giving the right answer, which is 2.

Step by step solution

01

Evaluate Original Limit Expression (Part a)

First, substitute the expression \( \lim \limits_{x \rightarrow 0} \frac{x^2-2x}{x^2-\sin x} \) directly to verify continuity or indeterminate form. Substitution gives \( \frac{0^2-2 \times 0}{0^2 - \sin 0} = \frac{0}{0} \), indicating an indeterminate form. Hence, we must use L'H么pital's Rule to resolve it.
02

Apply L'H么pital's Rule (Part a)

According to L'H么pital's Rule, if \( \lim \limits_{x \rightarrow a} \frac{f(x)}{g(x)} \) results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then we can differentiate the numerator and the denominator until the limit can be direct calculated. Differentiate the numerator: \( f'(x) = 2x-2 \) and the denominator: \( g'(x) = 2x-\cos x \). Recalculate the limit: \( \lim \limits_{x \rightarrow 0} \frac{2x-2}{2x-\cos x} \). Substituting \( x = 0 \), we get \( \frac{-2}{-1} = 2 \). The answer is 2, indicating an error in the original solution presented as "1".
03

Validate and Evaluate (Part b)

In Part b, we start with \( \lim \limits_{x \rightarrow 0} \frac{x^2-2x}{x^2-\sin x} \), which we know simplifies as done in Step 2 by L'H么pital's Rule. The given solution directly concludes \( \frac{-2}{-1} = 2 \) without seeming calculation error. Thus, this part is appropriately arriving at the correct solution, contradicting Part a's final result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indeterminate Forms
When evaluating limits, especially as functions approach a certain point, we often encounter what are called "indeterminate forms." These occur when substituting the limit value into a function results in expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These expressions don't give a clear answer because they suggest multiple potential outcomes.
Indeterminate forms require techniques beyond simple substitution to resolve, such as L'H么pital's Rule or algebraic manipulation. By recognizing these forms, we can proceed with appropriate methods to find meaningful limits.
Limit Evaluation
Limit evaluation is a fundamental concept in calculus, used to determine the value that a function approaches as the input approaches a specific point. It tests how a function behaves near that point rather than exactly at it鈥攅specially useful for functions not defined at certain values.
When limits result in indeterminate forms, as seen with \( \lim \limits_{x \rightarrow 0} \frac{x^2-2x}{x^2-\sin x} \), direct substitution isn't enough. Instead, evaluating such limits often involves simplifying the function or applying calculus techniques, like differentiation, to find precise values.
Differentiation
Differentiation is a process in calculus used to find the rate at which something changes. Applied to functions, it provides derivatives, which are fundamental in applying L'H么pital's Rule to resolve indeterminate forms.
In the provided example, when \( \lim \limits_{x \rightarrow 0} \frac{x^2-2x}{x^2-\sin x} \) resulted in \( \frac{0}{0} \), differentiation transformed the limit expression into \( \lim \limits_{x \rightarrow 0} \frac{2x-2}{2x-\cos x} \). By differentiating the numerator and denominator separately, we simplified the indeterminate form into a regular fraction, leading to a clear evaluation of the limit.
Continuity in Calculus
Continuity in calculus refers to a function's property of having no breaks, jumps, or gaps at a point or within an interval. If a function is continuous at a point, the limit as it approaches that point equals the function's value at that point.
In limits, continuity often helps determine if direct substitution will give a result. However, for functions not continuous at a point, like \( \frac{x^2-2x}{x^2-\sin x} \) at \( x = 0 \), we encounter indeterminate forms, necessitating alternative techniques like L'H么pital's Rule, to establish the limit's true value.

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Most popular questions from this chapter

Show that \(e^{x}\) grows faster as \(x \rightarrow \infty\) than \(x^{n}\) for any positive integer \(n,\) even \(x^{1,000,000} .\) (Hint: What is the \(n\) th derivative of \(x^{n} ?\) )

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