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Chapter 7: Problem 5

Use \(I^{\prime}\) H么pital's Rule to evaluate the limit. Then evaluate the limit using a method studied in Chapter 2. $$\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}$$

Short Answer

Expert verified
The limit is \( \frac{1}{2} \).

Step by step solution

01

Identify the Indeterminate Form

First, plug in the value of the limit into the function to check if it's in an indeterminate form. Evaluate \ \( \lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}} \) by substituting \( x = 0 \). This results in \( \frac{1 - \cos(0)}{0^2} = \frac{0}{0} \), indicating an indeterminate form.
02

Apply L'H么pital's Rule

Since the limit is in the form \(\frac{0}{0}\), we apply L'H么pital's Rule, which states that \( \lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)} \) given that the limit of the derivatives exists. Differentiate the numerator and the denominator separately. \( f(x) = 1 - \cos x \) gives \( f'(x) = \sin x \). \( g(x) = x^2 \) gives \( g'(x) = 2x \). Apply the derivatives to get a new limit expression: \( \lim_{x \to 0} \frac{\sin x}{2x} \).
03

Evaluate the New Limit Expression

Now simplify and evaluate the limit \( \lim_{x \to 0} \frac{\sin x}{2x} \). We know from trigonometric limits that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). Therefore, \( \lim_{x \to 0} \frac{\sin x}{2x} = \frac{1}{2} \).
04

Verify with Taylor Series Expansion

To verify, use a Taylor Series expansion of \( \cos x \) near zero: \( \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \cdots \). Substituting in the limit expression we get: \[ 1 - \cos x = \frac{x^2}{2} - \frac{x^4}{4!} + \cdots \]. Thus, the expression becomes \[ \lim_{x \rightarrow 0} \frac{\frac{x^2}{2} - \cdots}{x^2} = \lim_{x \rightarrow 0} \left( \frac{1}{2} - \cdots \right) = \frac{1}{2} \].

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Key Concepts

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

Indeterminate Forms
Indeterminate forms are expressions where it's not immediately clear what a limit is as a particular variable approaches a certain value. Imagine plugging zero into any expression that results in fractions like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or other combinations where direct substitution leads to undefined or ambiguous outcomes. These forms signal that we should proceed with caution.

In the exercise provided, the limit results in the indeterminate form \( \frac{0}{0} \). This means substituting the value directly doesn't work and requires additional techniques to find the actual limit. Recognizing this early on helps guide which mathematical tools鈥攍ike L'H么pital's Rule鈥攚e can use to solve the problem.
Trigonometric Limits
Trigonometric limits are a special type of limit that involve trigonometric functions such as sine, cosine, and tangent. Understanding these can significantly help in calculus. One of the most well-known trigonometric limits is \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). This particular limit often appears in problems involving trigonometric functions and serves as a fundamental building block for more complex limit evaluation using trig functions.

In our original exercise, after applying L'H么pital's Rule, the problem simplifies into a trigonometric limit: \( \lim_{x \to 0} \frac{\sin x}{2x} \). Knowing the limit of \( \frac{\sin x}{x} \), we can quickly conclude that this limit equals \( \frac{1}{2} \), simplifying the process significantly.
Taylor Series Expansion
The Taylor series expansion is a powerful technique that transforms functions into infinite sums of terms. It allows us to approximate functions by polynomials and is especially useful near certain points. For example, the Taylor series for \( \cos x \) around zero is \( \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \cdots \).

In the problem at hand, using a Taylor series expansion for \( \cos x \) lets us re-evaluate the original limit by substituting the series into the expression. This approach transforms the function into a manageable polynomial form, leading us to see the terms cancel in a way that reveals the limit is also \( \frac{1}{2} \). This provides a strong alternative verification to using calculus techniques like L'H么pital's Rule alone.

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Most popular questions from this chapter

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=(\sqrt{t})^{t}$$

Evaluate the integrals in Exercises \(107-110.\) $$\frac{1}{\ln a} \int_{1}^{x} \frac{1}{t} d t, \quad x>0$$

a. Graph \(f(x)=(\sin x)^{x}\) on the interval \(0 \leq x \leq \pi .\) What value would you assign to \(f\) to make it continuous at \(x=0 ?\) b. Verify your conclusion in part (a) by finding \(\lim _{x \rightarrow 0^{+}} f(x)\) with l'H么pital's Rule. c. Returning to the graph, estimate the maximum value of \(f\) on \([0, \pi] .\) About where is max \(f\) taken on? d. Sharpen your estimate in part (c) by graphing \(f^{\prime}\) in the same window to see where its graph crosses the \(x\) -axis. To simplify your work, you might want to delete the exponential factor from the expression for \(f^{\prime}\) and graph just the factor that has a zero.

Suppose that a cup of soup cooled from \(90^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) after \(10 \mathrm{min}\) in a room whose temperature was \(20^{\circ} \mathrm{C}\). Use Newton's Law of Cooling to answer the following questions. a. How much longer would it take the soup to cool to \(35^{\circ} \mathrm{C} ?\) b. Instead of being left to stand in the room, the cup of \(90^{\circ} \mathrm{C}\) soup is put in a freezer whose temperature is \(-15^{\circ} \mathrm{C}\). How long will it take the soup to cool from \(90^{\circ} \mathrm{C}\) to \(35^{\circ} \mathrm{C} ?\)

Find the domain and range of each composite function. Then graph the composites on separate screens. Do the graphs make sense in each case? Give reasons for your answers. Comment on any differences you see. a. \(y=\sin ^{-1}(\sin x)\) b. \(y=\sin \left(\sin ^{-1} x\right)\)
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