/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}

91Ó°ÊÓ

Find the limits in Exercises 21–36. $$ \lim _{h \rightarrow 0} \frac{\sin (\sin h)}{\sin h} $$

Short Answer

Expert verified
The limit is 1.

Step by step solution

01

Evaluate when h is 0

To find the limit as \(h\) approaches 0, we check what happens when \(h = 0\). Since \(\sin(0) = 0\), substituting directly gives \(\frac{\sin(\sin(0))}{\sin(0)} = \frac{0}{0}\), an indeterminate form.
02

Apply the limit identity

We use the identity \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \).
03

Substitute sine of sine

Notice that as \(h\to 0\), \(\sin(h)\to h\). Hence, by applying the identity twice: \[ \lim_{h \to 0} \frac{\sin(\sin(h))}{\sin(h)} = \lim_{u \to 0} \frac{\sin(u)}{u} = 1, \] where \(u = \sin(h)\) and as \(h\to 0\), \(u\to 0\).
04

Conclusion of the limit

Since we have just established that \( \frac{\sin(\sin(h))}{\sin h} \approx \frac{\sin(u)}{u} \), where \(u\to 0\) as \(h \to 0\), the limit evaluates to 1 as per the standard limit identity.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Trigonometric Limits
Trigonometric limits involve evaluating the limits of functions that contain trigonometric components such as sine, cosine, or tangent as variables approach a particular value. These limits are crucial in calculus as they help in understanding the behavior of trigonometric functions near certain points, often around zero. The limit we are exploring is particularly interested in the behavior of the sine function as its input goes to zero. The distinct characteristic of trigonometric limits is that they often involve simplifications or transforms using identities or approximations to resolve limits like these. Here, the limit \( \lim_{h \rightarrow 0} \frac{\sin(\sin h)}{\sin h} \)is challenging because direct substitution results in an indeterminate form, which leads us to use specific techniques such as trigonometric identities to make the limit solvable.
Indeterminate Forms
Indeterminate forms occur when substituting directly into a limit results in expressions like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or \( 0 \times \infty \). These suggest ambiguous values that do not lead directly to a number. In this exercise, substituting \( h = 0 \) yields the expression \( \frac{0}{0} \). This form requires other algebraic or analytical strategies to evaluate the limit. Techniques for solving indeterminate forms include:
  • Using L'Hôpital's Rule: Differentiating the top and bottom.
  • Simplifying the expression: Factoring or canceling out terms.
  • Utilizing standard limits or identities: Applying known results such as the limit identity for sine.
By recognizing the form as \( \frac{0}{0} \), we apply trigonometric limit identities to resolve the expression accurately.
Sine Function
The sine function, denoted as \( \sin(x) \), is a fundamental trigonometric function. It describes the y-coordinate of a point on the unit circle as the angle \( x \) varies. This function is periodic, smooth, and continuous, which are helpful properties when dealing with limits. The key feature of the sine function relevant to limits is its behavior near zero, where \( \sin(x) \approx x \). This approximation is useful when addressing limits involving the sine function, particularly since its deviation around zero is minor enough that we can utilize the identity \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1. \)For our problem, this approximation helps us express complex trigonometric limits in a more straightforward form.
Limit Identity
A limit identity is a well-established result used to determine the limit of an expression without performing direct substitution. One of the most critical limit identities in calculus involving trigonometric functions is:\( \lim_{x \to 0} \frac{\sin(x)}{x} = 1.\)This identity indicates that as \( x \) becomes very small, the fraction \( \frac{\sin(x)}{x} \) approximates 1. It simplifies many problems involving trigonometric functions near zero. In our example, when evaluating the limit \( \lim_{h \rightarrow 0} \frac{\sin(\sin h)}{\sin h},\)we substitute \( u = \sin(h) \), rendering the expression to a form where the identity can be applied, thereby allowing us to conclude the limit is 1. Such identities prove indispensable in calculus, particularly when tackling seemingly complex problems efficiently.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises \(43-46\) , find a function that satisfies the given conditions and sketch its graph. (The answers here are are not unique. Any function that satisfies the conditions is acceptable. Feel free to use formulas defined in pieces if that will help.) $$ \lim _{x \rightarrow \pm \infty} f(x)=0, \lim _{x \rightarrow 2^{-}} f(x)=\infty, \text { and } \lim _{x \rightarrow 2^{+}} f(x)=\infty $$

Sometimes a change of variable can change an unfamiliar expression into one whose limit we know how to find. For example, $$ \begin{aligned} \lim _{x \rightarrow \infty} \sin \frac{1}{x} &=\lim _{\theta \rightarrow 0^{+}} \sin \theta \\ &=0 \end{aligned} $$ This suggests a creative way to "see" limits at infinity. Describe the procedure and use it to picture and determine limits in Exercises \(79-84 .\) $$ \lim _{x \rightarrow \pm \infty} x \sin \frac{1}{x} $$

Suppose that \(f(x)\) and \(g(x)\) are polynomials in \(x\) . Can the graph of \(f(x) / g(x)\) have an asymptote if \(g(x)\) is never zero? Give reasons for your answer.

In Exercises \(43-46\) , find a function that satisfies the given conditions and sketch its graph. (The answers here are are not unique. Any function that satisfies the conditions is acceptable. Feel free to use formulas defined in pieces if that will help.) $$ \lim _{x \rightarrow \pm \infty} g(x)=0, \lim _{x \rightarrow 3^{-}} g(x)=-\infty, \text { and } \lim _{x \rightarrow 3^{+}} g(x)=\infty $$

Find the limits in Exercises \(29-34 .\) Are the functions continuous at the point being approached? $$ \lim _{t \rightarrow 0} \sin \left(\frac{\pi}{2} \cos (\tan t)\right) $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.