Question

, use a calculator to find the indicated limit. Use a graphing calculator to plot the function near the limit point. $$ \lim _{x \rightarrow 0} \frac{(x-\sin x)^{2}}{x^{2}} $$

Short answer

The limit is 0.

Step-by-step solution

Analyze the Function

We start by analyzing the function \( \frac{(x - \sin x)^2}{x^2} \). Notice that both the numerator \((x - \sin x)^2\) and the denominator \(x^2\) approach 0 as \(x\) approaches 0. This indicates that the limit is in indeterminate form \(\frac{0}{0}\).

Use L'Hôpital's Rule

Since we have an indeterminate form \(\frac{0}{0}\), we can apply L'Hôpital's Rule. However, before doing so, let's use successive approximations by graphing and evaluating numerically around \(x = 0\).

Plot the Function

Graph the function \(y = \frac{(x - \sin x)^2}{x^2}\) near \(x = 0\) using a graphing calculator. Set the x-axis to include negative and positive values close to 0, such as from -0.1 to 0.1. Observe the behavior of the function's graph close to \(x = 0\). You should see that the graph flattens towards a certain value as \(x\) approaches 0.

Calculate Numeric Approximations

Using the calculator, compute the function's values for \(x\) values close to 0, like 0.01, -0.01, 0.001, and -0.001. The values approach a number. This step confirms the behavior seen in the graph.

Confirm Result with L'Hôpital's Rule

Differentiate the numerator: the derivative of \((x - \sin x)^2\) with respect to \(x\) is \(2(x - \sin x)(1 - \cos x)\). The derivative of the denominator \(x^2\) is \(2x\). So, apply L'Hôpital's rule: \[\lim_{x \to 0} \frac{2(x - \sin x)(1 - \cos x)}{2x}\]This simplifies to \[\lim_{x \to 0} \frac{(x - \sin x)(1 - \cos x)}{x}\].

Simplify Further Using Series Expansion

Look at the Taylor series expansion for \(\sin x \approx x - \frac{x^3}{6}\) and \(1 - \cos x \approx \frac{x^2}{2}\). Substitute these approximations:\[(x - \sin x) \approx \frac{x^3}{6}\]and \[1 - \cos x \approx \frac{x^2}{2}\]. Substitute back to get:\[\lim_{x \to 0} \frac{\left(\frac{x^3}{6}\right) \cdot \left(\frac{x^2}{2}\right)}{x} = \lim_{x \to 0} \frac{x^5}{12x} = \lim_{x \to 0} \frac{x^4}{12} = 0\].

Conclusion

From the numerical evaluation, graphical interpretation, and calculus evaluation, the limit of the function as \(x\) approaches 0 is confirmed to be 0.

Key concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to find limits that result in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). In these cases, direct substitution doesn't work, and that's when L'Hôpital's Rule becomes handy. It states that if the limit of \( \frac{f(x)}{g(x)} \) as \( x \) approaches a point results in an indeterminate form, then the limit is the same as the limit of \( \frac{f'(x)}{g'(x)} \), provided this limit exists. This involves differentiating both the numerator and the denominator separately until you get a determinate form.

For instance, in our problem of \( \lim_{x \to 0} \frac{(x - \sin x)^2}{x^2} \), upon finding it is \( \frac{0}{0} \), we differentiate to use L'Hôpital's Rule:
\[ f'(x) = 2(x - \sin x)(1 - \cos x) \] and \[ g'(x) = 2x \]. This allows simplifying the expression further, bringing us closer to finding the limit.

Indeterminate Forms

Indeterminate forms are expressions that do not lead directly to a particular value when trying to evaluate a limit. Common examples include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), and others like \( 0 \cdot \infty \) or \( \infty - \infty \). These forms suggest more work is needed to find the actual limit.

In our given problem \( \lim_{x \to 0} \frac{(x - \sin x)^2}{x^2} \), as \( x \) approaches 0, both \( (x - \sin x)^2 \) and \( x^2 \) approach 0, creating an indeterminate form \( \frac{0}{0} \). This requires us to use techniques such as L'Hôpital's Rule or series expansion, to resolve the expression and find the limit. These scenarios can often unravel deeper behaviors of functions near points of interest.

Taylor Series Expansion

Taylor Series Expansion is a method in calculus where functions are expressed as an infinite sum of terms calculated from the values of their derivatives at a single point. This process is invaluable for approximating complex functions. It's particularly useful near the center of expansion and provides a way to approximate nonlinear functions with polynomials.

In resolving our function \( \lim_{x \to 0} \frac{(x - \sin x)^2}{x^2} \), we utilized the Taylor series to represent \( \sin x \) and \( \cos x \):

• \( \sin x \approx x - \frac{x^3}{6} \)
• \( 1 - \cos x \approx \frac{x^2}{2} \)

By substituting these approximations into the limit expression, it becomes easier to simplify and evaluate the behavior of the function as \( x \) approaches 0. This makes Taylor series a crucial technique in handling indeterminate forms without explicitly computing derivatives repeatedly.