/*! 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 30 Find the limit of the following ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the limit of the following sequences or determine that the limit does not exist. $$\\{n(1-\cos (1 / n))\\}$$

Short Answer

Expert verified
Answer: The limit of the sequence is 0.

Step by step solution

01

Rewrite the original expression

First, let's rewrite the original expression, \(a_n = n(1-\cos (1/n))\), such that it would be clearer to use L'Hôpital's rule. Define a new variable \(x=1/n\). The expression becomes: $$a_n = \frac{1-\cos x}{x}$$ Now, we want to find the limit of \(a_n\) as \(n\) approaches infinity, which is equivalent to finding the limit of the above expression as \(x\) approaches \(0\).
02

Apply L'Hôpital's Rule

Since the limit has the indeterminate form \(\frac{0}{0}\) as \(x\) approaches \(0\), we can apply L'Hôpital's rule: 1. Differentiate the numerator and denominator with respect to \(x\). 2. Check if the new limit exists. If it does, we have found the limit; otherwise, repeat step 1. First, derivative of the numerator and denominator: $$\frac{d}{dx}(1-\cos x) = \sin x$$ $$\frac{d}{dx}(x) = 1$$ Now let's compute the limit: $$\lim_{x\to 0} \frac{\sin x}{1} = \lim_{x\to 0} \sin x$$
03

Evaluate the limit

As the limit of \(\frac{d}{dx}(1-\cos(x))\) divided by the limit of \(\frac{d}{dx}(x)\) exists and is equal to: $$\lim_{x\to 0}\sin x = \sin 0 = 0$$ We have found that: $$\lim_{n\to \infty}\left\{n\left(1-\cos\left(\frac{1}{n}\right)\right)\right\}=\lim_{x\to 0}\frac{1-\cos x}{x}=0$$ So, the limit of the given sequence exists and is equal to \(0\).

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Key Concepts

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

L'Hôpital's Rule
L'Hôpital's Rule is a powerful tool in calculus for finding limits, especially when dealing with indeterminate forms like \(0/0\) or \( \infty/\infty \). It allows us to transform complex limit problems into simpler ones by differentiating the numerator and the denominator until a determinate form is achieved. This method is straightforward but requires that both the original function's numerator and denominator can be differentiated. The essence of L'Hôpital's Rule is:
  • Identify the indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
  • Differentiates the numerator and the denominator separately.
  • Recompute the limit of the new function resulting from the derivatives.
  • If the new limit remains indeterminate, repeat the process.
In our exercise, we rearranged the sequence \( a_n = n(1 - \cos(1/n)) \) into the function \( f(x) = \frac{1 - \cos x}{x} \), when expressed in terms of \( x \). By applying L'Hôpital’s Rule, we differentiate and simplify our function to obtain a definitive limit of zero. This systematic approach makes it easier to tackle problems involving indeterminate forms.
Indeterminate Forms
Indeterminate forms occur when a limit is not immediately clear due to an undefined mathematical structure, like \(0/0\) or \(\infty/\infty\). These forms need special techniques to resolve, because conventional substitution in limits leads to ambiguity. Indeterminate forms signal that more work, like applying derivatives, is needed to find a limit.
In our sequence problem, when transformed to \(\frac{1-\cos x}{x}\), the limit \(\lim_{x \to 0} \) hits an indeterminate form \(\frac{0}{0}\).
This requires using L'Hôpital's Rule to differentiate the numerator and the denominator separately before recalculating the limit. Resolving such forms is crucial to simplifying expressions and verifying the existence of limits, as demonstrated in the process of reaching a limit of zero for our sequence.
Cosine Function
The cosine function, a fundamental trigonometric function, is periodic and oscillates between -1 and 1. It includes important properties that influence limit calculation, such as:
  • \( \cos(0) = 1 \)
  • \(\cos(x)\) is even, satisfying \(\cos(-x) = \cos(x)\)
These properties are useful when dealing with limits and derivatives involving the cosine function.
In the context of our exercise, understanding cosine's behavior near zero is crucial. The small-angle approximation indicates \( 1 - \cos(x) \approx \frac{x^2}{2} \) for small values of \( x \), aiding in the simplification and differentiation needed to resolve indeterminate forms. This approximation, however, was not directly used here because L'Hôpital's Rule easily resolves the expression after determining the derivative of the cosine function \( \sin x \). Understanding these properties about cosine and how they behave in derivative form is vital when such functions are part of a limit problem.

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Most popular questions from this chapter

Consider the geometric series \(f(r)=\sum_{k=0}^{\infty} r^{k},\) where \(|r|<1\) a. Fill in the following table that shows the value of the series \(f(r)\) for various values of \(r\) $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline r & -0.9 & -0.7 & -0.5 & -0.2 & 0 & 0.2 & 0.5 & 0.7 & 0.9 \\ \hline f(r) & & & & & & & & & \\ \hline \end{array}$$ b. Graph \(f,\) for \(|r|<1\) \text { c. Evaluate } \lim _{r \rightarrow 1^{-}} f(r) \text { and } \lim _{r \rightarrow-1^{+}} f(r)

a. Sketch the function \(f(x)=1 / x\) on the interval \([1, n+1]\) where \(n\) is a positive integer. Use this graph to verify that $$\ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}<1+\ln n.$$ b. Let \(S_{n}\) be the sum of the first \(n\) terms of the harmonic series, so part (a) says \(\ln (n+1)0,\) for \(n=1,2,3, \ldots\) c. Using a figure similar to that used in part (a), show that $$\frac{1}{n+1}>\ln (n+2)-\ln (n+1).$$ d. Use parts (a) and (c) to show that \(\left\\{E_{n}\right\\}\) is an increasing sequence \(\left(E_{n+1}>E_{n}\right)\). e. Use part (a) to show that \(\left\\{E_{n}\right\\}\) is bounded above by 1 . f. Conclude from parts (d) and (e) that \(\left\\{E_{n}\right\\}\) has a limit less than or equal to \(1 .\) This limit is known as Euler's constant and is denoted \(\gamma\) (the Greek lowercase gamma). g. By computing terms of \(\left\\{E_{n}\right\\}\), estimate the value of \(\gamma\) and compare it to the value \(\gamma \approx 0.5772 .\) (It has been conjectured that \(\gamma\) is irrational.) h. The preceding arguments show that the sum of the first \(n\) terms of the harmonic series satisfy \(S_{n} \approx 0.5772+\ln (n+1)\) How many terms must be summed for the sum to exceed \(10 ?\)

Marie takes out a \(\$ 20,000\) loan for a new car. The loan has an annual interest rate of \(6 \%\) or, equivalently, a monthly interest rate of \(0.5 \% .\) Each month, the bank adds interest to the loan balance (the interest is always \(0.5 \%\) of the current balance), and then Marie makes a \(\$ 200\) payment to reduce the loan balance. Let \(B_{n}\) be the loan balance immediately after the \(n\) th payment, where \(B_{0}=\$ 20,000\). a. Write the first five terms of the sequence \(\left\\{B_{n}\right\\}\). b. Find a recurrence relation that generates the sequence \(\left\\{B_{n}\right\\}\). c. Determine how many months are needed to reduce the loan balance to zero.

Use Theorem 8.6 to find the limit of the following sequences or state that they diverge. $$\left\\{\frac{n^{1000}}{2^{n}}\right\\}$$

In the following exercises, two sequences are given, one of which initially has smaller values, but eventually "overtakes" the other sequence. Find the sequence with the larger growth rate and the value of \(n\) at which it overtakes the other sequence. $$a_{n}=n^{10} \text { and } b_{n}=n^{9} \ln ^{3} n, n \geq 7$$

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