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Use the Squeeze Theorem to prove that \(\lim _{x \rightarrow 0^{+}}\left[\sqrt{x} \cos ^{2}(1 / x)\right]=0\) Identify the functions \(f\) and \(h,\) show graphically that \(f(x) \leq \sqrt{x} \cos ^{2}(1 / x) \leq h(x)\) for all \(x > 0,\) and justify \(\lim _{x \rightarrow 0^{+}} f(x)=0\) and \(\lim _{x \rightarrow 0^{+}} h(x)=0\)

Short Answer

Expert verified
By applying the Squeeze Theorem, it was proven that \( \lim_{x \rightarrow 0^{+}} \sqrt{x}\cos^{2}(1/x) = 0 \).

Step by step solution

01

Identify functions f(x) and g(x)

Let's choose f(x) = 0 and g(x) = \(\sqrt{x}\) as our lower and upper bounds respectively. It can be observed that for \(x > 0\), the function values of \(\sqrt{x} \cos^{2}(1/x)\) must always lie in between 0 and \(\sqrt{x}\), because \( 0 \le \cos^{2}(1/x) \le 1 \) for any x values, and when multiplied by \(\sqrt{x}\), it's guaranteed that \(0 \le \sqrt{x} \cos^{2}(1/x) \le \sqrt{x}\).
02

Visualize and confirm the functions

If you would sketch the functions \(f(x) = 0\), \(g(x) = \sqrt{x}\) and \(h(x) = \sqrt{x}\cos^{2}(1/x)\), you would see that 'h'(the middle function), sits comfortably between 'f'(the bottom function) and 'g'(the top function). That means \(f(x) \le h(x) \le g(x)\).
03

Compute limits of f(x) and g(x)

Use the fact that as \(x \rightarrow 0^{+}, f(x)=0\). Also remember that as \(x \rightarrow 0^{+}, g(x)=\sqrt{x}\) also approaches zero. Thus \(\lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} g(x) = 0 \).
04

Apply Squeeze Theorem

From the Squeeze Theorem, since \(\lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} g(x) = 0 \) and \(f(x) \le h(x) \le g(x)\), it follows that \(\lim_{x \rightarrow 0^{+}} h(x) = 0 \).

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Key Concepts

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

Limits of Functions
Understanding the concept of limits is critical in calculus, as it helps grasp behaviors of functions as they approach a certain point. A limit evaluates the value that a function is approaching as the input approaches some value. For example, in our exercise, we look at the limit of \( \sqrt{x} \cos^2(1/x) \) as \( x \) approaches 0 from the positive side (\( x \to 0^+ \) ).

The limit can often be straightforward, but in cases where direct substitution yields an indeterminate form, like 0/0, or, as in our context, where the behavior is oscillatory or non-standard, we may use limit laws or theorems such as the Squeeze Theorem. This particular theorem is useful when the function we're evaluating is 'squeezed' between two other functions whose limits we know, and we understand that our function cannot exceed those bounds.
Continuity in Calculus
Continuity of a function at a point means the function's behavior is predictable and uninterrupted at that point. More technically, a function \( f \) is continuous at a point \( a \) if \( \lim_{x \to a} f(x) = f(a) \). In the context of our exercise, knowing about continuity is helpful when working with limits, as continuous functions make limit computations straightforward.

However, for functions that are not easily defined at a point, like \( \sqrt{x} \cos^2(1/x) \) approaching 0, we can't directly observe continuity. That's where the Squeeze Theorem lends a hand. If we know the limiting behavior of a function's bounds and these bounds happen to be continuous, which is the case with \( f(x) = 0 \) and \( g(x) = \sqrt{x} \), and these bounds converge to the same value at a point, the function squeezed between them shares the same limit at that point, revealing a sort of 'indirect continuity'.
Trigonometric Functions
Trigonometric functions are fundamental in mathematics, involving sine, cosine, tangent, and their reciprocals. They oscillate between specific values, with the cosine function, for example, varying between -1 and 1. In our exercise, \( \cos^2(1/x) \) is a trigonometric function with the square causing all results to be non-negative, fluctuating between 0 and 1.

As \( x \) gets closer to 0, \( \cos^2(1/x) \) oscillates more rapidly since \(1/x \) grows without bound. These oscillations could make it difficult to determine the limit directly. However, irrespective of the oscillations, the range of \( \cos^2(1/x) \) allows the function \( \sqrt{x} \cos^2(1/x) \) to stay within the bounds of \( 0 \) and \( \sqrt{x} \), assisting us in applying the Squeeze Theorem. Understanding the predictable behavior of trigonometric functions, even in complex scenarios, improves our limit-finding toolkit.

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Most popular questions from this chapter

Suppose you ease your car up to a stop sign at the top of a hill. Your car rolls back a couple of feet and then you drive through the intersection. A police officer pulls you over for not coming to a complete stop. Use the Intermediate Value Theorem to argue that there was an instant in time when your car was stopped (in fact, there were at least two). What is the difference between this stopping and the stopping that the police officer wanted to see?

Use numerical and graphical evidence to conjecture whether the limit at \(x=a\) exists. If not, describe what is happening at \(x=a\) graphically. $$\lim _{x \rightarrow 0} \sin \left(\frac{1}{x}\right)$$

Prove that the limit is correct using the appropriate definition (assume that \(k\) is an integer). $$\lim _{x \rightarrow 5} \frac{4}{(x-5)^{2}}=\infty$$

A function is continuous from the right at \(x=a\) if \(\lim _{x \rightarrow a^{+}} f(x)=f(a) .\)Determine whether \(f(x)\) is continuous from the right at \(x=2.\) $$f(x)=\left\\{\begin{array}{ll} x^{2} & \text { if } x<2 \\ 3 x-2 & \text { if } x>2 \end{array}\right.$$

In this exercise, we explore the definition of \(\lim _{x \rightarrow 2} x^{2}=4\) with \(\varepsilon=0.1 .\) Show that \(x^{2}-4 < 0.1\) if \(2 < x < \sqrt{4.1}\). This indicates that \(\delta_{1}=0.02484\) works for \(x > 2 .\) Show that \(x^{2}-4 > -0.1\) if \(\sqrt{3.9} < x < 2 .\) This indicates that \(\delta_{2}=0.02515\) works for \(x < 2 .\) For the limit definition, is \(\delta=\delta_{1}\) or \(\delta=\delta_{2}\) the correct choice? Briefly explain.

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