/*! 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} Problem 9 Find the limits in Exercises \(9... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the limits in Exercises \(9-12\) . $$\lim _{x \rightarrow \infty} \frac{\sin 2 x}{x}$$

Short Answer

Expert verified
The limit is 0.

Step by step solution

01

Analyze the Limit Expression

We are tasked to find the limit as \( x \) approaches infinity for the function \( \frac{\sin 2x}{x} \). First, note that \( \sin 2x \) oscillates between -1 and 1 for all values of \( x \). This means that \( -1 \leq \sin 2x \leq 1 \).
02

Apply the Squeeze Theorem

Given that \( -1 \leq \sin 2x \leq 1 \), we can divide the entire inequality by \( x \) (where \( x > 0 \)) to get \( -\frac{1}{x} \leq \frac{\sin 2x}{x} \leq \frac{1}{x} \).
03

Evaluate Limits of Bounds

Consider the limits of the bounds as \( x \rightarrow \infty \): - \( \lim_{x \to \infty} \left(-\frac{1}{x}\right) = 0 \) - \( \lim_{x \to \infty} \left(\frac{1}{x}\right) = 0 \)
04

Conclude Using the Squeeze Theorem

Since \( -\frac{1}{x} \leq \frac{\sin 2x}{x} \leq \frac{1}{x} \) and both bounds approach 0 as \( x \rightarrow \infty \), by the Squeeze Theorem, the limit \( \lim_{x \to \infty} \frac{\sin 2x}{x} \) must also be 0.

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Key Concepts

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

Squeeze Theorem
The Squeeze Theorem is a powerful tool in calculus. It helps us find the limit of a function when the direct evaluation might be challenging.
To use the Squeeze Theorem, we need three functions: two that "squeeze" the function of interest.
Essentially, if we have a function, say \( f(x) \), and we know that it is always greater than another function \( g(x) \) and less than \( h(x) \), and both \( g(x) \) and \( h(x) \) approach the same limit as \( x \) approaches a particular value, then \( f(x) \) must approach that same limit.
  • If \( g(x) \leq f(x) \leq h(x) \) for all \( x \), and both \( \lim_{x \to a} g(x) = L \) and \( \lim_{x \to a} h(x) = L \), then \( \lim_{x \to a} f(x) = L \) as well.
In our example, the function \( \frac{\sin 2x}{x} \) is squeezed between \( -\frac{1}{x} \) and \( \frac{1}{x} \). As both these functions approach zero as \( x \) goes to infinity, so does \( \frac{\sin 2x}{x} \).
This makes the Squeeze Theorem a very effective method for proving the limits of functions that might be sandwiched between two simpler limits.
Infinity Limits
Infinity limits deal with the behavior of functions as the input, often denoted as \( x \), grows larger without bound or approaches negative infinity. These limits help us understand how functions behave in extreme cases.
Calculating limits as \( x \to \infty \) can help in predicting long-term behavior in models and scenarios such as population growth or economic trends.
  • These limits are common in many mathematical models where variables increase without end.
  • When evaluating infinity limits, look to simplify functions and expressions using known limits or approximations.
In our particular exercise, we're considering \( \lim_{x \to \infty} \frac{\sin 2x}{x} \).
Here, \( \sin 2x \) keeps oscillating between 1 and -1 no matter how big \( x \) gets, but the denominator \( x \) continues to grow. Therefore, our function's overall magnitude shrinks, tending to zero as \( x \) becomes very large.
Trigonometric Functions
Trigonometric functions are fundamental in both pure and applied mathematics. They describe relationships in triangles and model periodic phenomena such as sound and light waves.
Key trigonometric functions include sine, cosine, and tangent, which relate the angles of a right triangle to the ratios of its sides.
The sine function, represented as \( \sin x \), has values ranging between -1 and 1.
It is periodic with a period of \( 2\pi \), meaning the function repeats its values in every interval of \( 2\pi \).
  • In calculus, \( \sin x \) and other trigonometric functions often appear in limits and integrals.
  • Their periodic nature makes them interesting for limit problems, especially when combined with non-periodic functions like polynomials.
In the example, \( \sin 2x \) oscillates between -1 and 1, but when divided by \( x \), it illustrates how its influence diminishes as \( x \) increases. Understanding this diminishing effect is crucial for evaluating the limit as \( x \to \infty \).

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Most popular questions from this chapter

If \(x^{4} \leq f(x) \leq x^{2}\) for \(x\) in \([-1,1]\) and \(x^{2} \leq f(x) \leq x^{4}\) for \(x<-1\) and \(x>1,\) at what points \(c\) do you automatically know \(\lim _{x \rightarrow c} f(x) ?\) What can you say about the value of the limit at these points?

Let \(g(x)=\sqrt{x}\) for \(x \geq 0.\) a. Find the average rate of change of \(g(x)\) with respect to \(x\) over the intervals \([1,2],[1,1.5]\) and \([1,1+h] .\) b. Make a table of values of the average rate of change of \(g\) with respect to \(x\) over the interval \([1,1+h]\) for some values of \(h\) approaching zero, say \(h=0.1,0.01,0.001,0.0001,0.00001,\) and \(0.000001 .\) c. What does your table indicate is the rate of change of \(g(x)\) with respect to \(x\) at \(x=1 ?\) d. Calculate the limit as \(h\) approaches zero of the average rate of change of \(g(x)\) with respect to \(x\) over the interval \([1,1+h].\)

If the product function \(h(x)=f(x) \cdot g(x)\) is continuous at \(x=0,\) must \(f(x)\) and \(g(x)\) be continuous at \(x=0 ?\) Give reasons for your answer.

In Exercises \(47-50,\) graph the function \(f\) to see whether it appears to have a continuous extension to the origin. If it does, use Trace and Zoom to find a good candidate for the extended function's value at \(x=0\) .If the function does not appear to have a continuous extension, can it be extended to be continuous at the origin from the right or from the left? If so, what do you think the extended function's value(s) should be? $$ f(x)=(1+2 x)^{1 / x} $$

In Exercises \(73-76,\) find a function that satisfies the given conditions and sketch its graph. (The answers here are not unique. Any function that satisfies the conditions is acceptable. Feel free to use formulas defined in pieces if that will help.) $$\lim _{x \rightarrow \pm \infty} k(x)=1, \lim _{x \rightarrow 1^{-}} k(x)=\infty, \text { and } \lim _{x \rightarrow 1^{+}} k(x)=-\infty$$

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