Problem 76 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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'Hôpital's Rule to resolve it.

Step 2: Apply L'Hôpital's Rule (Part a)

According to L'Hôpital'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".

Step 3: 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'Hôpital'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.

Key Concepts

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'Hôpital'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—especially 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'Hôpital'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'Hôpital's Rule, to establish the limit's true value.