Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow 0} \frac{2 x-\sin x}{x} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Check the Form

First, substitute \( x = 0 \) into the limit expression \( \frac{2x - \sin x}{x} \). This results in \( \frac{2(0) - \sin(0)}{0} = \frac{0}{0} \), which is an indeterminate form.

Step 2: Apply l'Hôpital's Rule

Since we have an indeterminate form \( \frac{0}{0} \), we can apply l'Hôpital's Rule. Differentiate the numerator and the denominator separately. The derivative of the numerator \( 2x - \sin x \) is \( 2 - \cos x \). The derivative of the denominator \( x \) is \( 1 \).

Step 3: Reevaluate the Limit

Now, consider the limit \( \lim_{x \to 0} \frac{2 - \cos x}{1} \). Substitute \( x = 0 \) to evaluate the limit: \( \frac{2 - \cos(0)}{1} = 2 - 1 = 1 \).

Step 4: State the Result

After using l'Hôpital's Rule and simplifying, the resulting limit is 1.

Key Concepts

Indeterminate Forms

When dealing with limits, indeterminate forms are expressions that do not initially present a clear value or result. A common indeterminate form is \( \frac{0}{0} \), which arises when both the numerator and denominator of a fraction approach zero. This form indicates a need to further analyze the expression to find a well-defined limit. In the problem \( \lim _{x \rightarrow 0} \frac{2x - \sin x}{x} \), substituting \( x = 0 \) results in \( \frac{0}{0} \). This tells us we need additional techniques, such as l'Hôpital's Rule, to properly evaluate the limit. Understanding and recognizing these forms is essential in calculus to apply suitable methods for resolving limits.

Limit Evaluation

Limit evaluation is the process of determining the value that a function approaches as the input approaches a certain point. In the exercise we have, limit evaluation starts with determining if the expression is in an indeterminate form, which it is. From there, we use advanced methods like l'Hôpital's Rule to evaluate the limit.

• First, substitute the point of interest (here, \( x = 0 \)) into the expression.
• Identify if the result is an indeterminate form (e.g., \( \frac{0}{0} \)).
• Use appropriate techniques to further evaluate the limit, often involving algebraic manipulation or calculus methods such as differentiation.

This careful procedure ensures that the final limit is correctly found, as illustrated when evaluating \( \lim_{x \to 0} \frac{2 - \cos x}{1} = 1 \) after applying l'Hôpital's Rule.

Differentiation

Differentiation is a calculus technique used to find the rate at which a function is changing at any given point. It involves finding the derivative of a function, which provides a slope of the tangent line at any point of interest. When an expression results in an indeterminate form like \( \frac{0}{0} \), one can use differentiation to evaluate the limit as shown in l'Hôpital's Rule.In our solution, we differentiate the numerator \( 2x - \sin x \) to get \( 2 - \cos x \) and the denominator \( x \) to get \( 1 \). Using these derivatives allows us to re-evaluate the original limit expression in a form that is no longer indeterminate. Differentiation is, therefore, a crucial step in turning a complex problem into a solvable one by providing new insights into how the function behaves at specific points.

Trigonometric Functions

Trigonometric functions like \( \sin x \) and \( \cos x \) are essential in calculus, especially when dealing with limits and differentiation. These functions describe relationships between the angles and sides of triangles, but they also feature prominently in various calculus problems, including evaluating limits.

• The sine function, \( \sin x \), returns values based on the circular path, starting from zero as \( x \) approaches 0.
• The cosine function, \( \cos x \), at \( x = 0 \) is 1, which is crucial in simplifying expressions after differentiation.

In the problem presented, the original expression involves \( \sin x \), which behaves predictably around zero, allowing the use of differentiation to simplify \( 2x - \sin x \) to \( 2 - \cos x \). Understanding these trigonometric behaviors can significantly simplify the process of evaluating limits through further calculus techniques.