Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

, use a calculator to find the indicated limit. Use a graphing calculator to plot the function near the limit point. $$ \lim _{x \rightarrow 3} \frac{x-\sin (x-3)-3}{x-3} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Understand the Limit Expression

The given limit expression is \( \lim _{x \rightarrow 3} \frac{x-\sin (x-3)-3}{x-3} \). This implies you need to find the value of the function \( f(x) = \frac{x-\sin (x-3)-3}{x-3} \) as \( x \) approaches 3.

Step 2: Simplify the Expression

Rewrite the function to see if it simplifies. The expression is \( f(x) = \frac{x- 3 - \sin(x-3)}{x-3} \). It takes the form \( \frac{0}{0} \), suggesting L'Hopital's Rule might be applicable.

Step 3: Apply L'Hopital's Rule

Use L'Hopital's Rule, which states that \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \), when \( \frac{0}{0} \) form occurs. Differentiate the numerator and the denominator:- Numerator: \( (x - \sin(x-3) - 3)' = 1 - \cos(x-3) \).- Denominator: \( (x - 3)' = 1 \).

Step 4: Evaluate the New Limit

Now calculate the limit: \( \lim_{x \to 3} \frac{1 - \cos(x-3)}{1} = 1 - \cos(3-3) = 1 - \cos(0) = 1 - 1 = 0 \). As \( x \to 3 \), the limit of \( \frac{1 - \cos(x-3)}{1} \) is 0.

Step 5: Plot the Function

Using a graphing calculator, plot the function \( f(x) = \frac{x- \sin(x-3)-3}{x-3} \) near \( x = 3 \). Observe the behavior of the function as \( x \) approaches 3. The graph should confirm that the function approaches a value of 0.

Key Concepts

Limits

Limits are fundamental in calculus, helping us understand the behavior of functions as they approach specific points. To evaluate a limit, you observe how the values of a function get closer to a particular value as the input approaches a given number. In this exercise, the limit involved is \( \lim _{x \rightarrow 3} \frac{x-\sin (x-3)-3}{x-3} \). This is asking, as \( x \) gets closer to 3, what value does the function approach?
When directly substituting the value \( x = 3 \) into the function, it leads to an indeterminate form \( \frac{0}{0} \). This indicates that more work is needed to find the limit. This initial point of indeterminacy often suggests that concepts such as algebraic manipulation or specific calculus rules can be applied to solve the problem effectively. Understanding limits is crucial because they form the foundation for derivatives and integrals, the core operations in calculus.

L'Hopital's Rule

L'Hopital's Rule is a method used to evaluate certain limits that aren't straightforward due to the indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It provides a way to find these limits by differentiating the numerator and denominator independently, then re-evaluating the limit.In our exercise, we encountered the indeterminate form \( \frac{0}{0} \) when substituting \( x = 3 \) into the function \( \frac{x-\sin (x-3)-3}{x-3} \). Hence, L'Hopital's Rule is applied:

• Differentiate the numerator: \( 1 - \cos(x-3) \).
• Differentiate the denominator: \( 1 \).

Now substitute \( x = 3 \) in the differentiated fraction: \( \frac{1 - \cos(3-3)}{1} = \frac{1 - \cos(0)}{1} = 1 - 1 = 0 \). This confirms the limit of the original function is 0 as \( x \) approaches 3. L'Hopital's Rule is a powerful tool when direct evaluation of limits leads to indeterminate forms, making complex limit problems much simpler to solve.

Trigonometric Limits

Trigonometric functions often appear in limit problems, calling for a comprehensive understanding of their properties and limit behavior. Basic trigonometric limits, such as \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) and \( \lim_{x \to 0} 1 - \cos x = 0 \), are crucial tools which can simplify and solve otherwise difficult problems.
Our problem involved trigonometric terms due to the \( \sin(x-3) \) function. When applying L'Hopital's Rule, knowing derivatives and limits of trigonometric functions is essential. The derivative of \( \sin(x-3) \) is \( \cos(x-3) \), which was essential in calculating the resulting limit. These foundations aid in evaluating limits by breaking down the functions into more familiar, simpler forms.
Understanding trigonometric limits helps in recognizing patterns and solutions, not just in theoretical calculus exercises but in practical applications that deal with waves, oscillations, and periodic phenomena.