/*! 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 21 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. $$\\{\sqrt{\left(1+\frac{1}{2 n}\right)^{n}}\\}$$

Short Answer

Expert verified
Answer: The limit of the given sequence as n approaches infinity is \(e^{\frac{1}{2}}\).

Step by step solution

01

Identify the sequence

The sequence is given as: $$\sqrt{\left(1+\frac{1}{2n}\right)^{n}}$$
02

Apply the limit

We need to find the limit of the sequence as n approaches infinity: $$\lim_{n \to \infty} \sqrt{\left(1+\frac{1}{2n}\right)^{n}}$$
03

Rewrite the sequence

To apply the exponential function limit, we will rewrite the sequence as: $$\lim_{n \to \infty} \left(1+\frac{1}{2n}\right)^{\frac{n}{2}}$$
04

Apply the property of the limiting form of the exponent

We know that: $$\lim_{x \to 0}(1+x)^{\frac{1}{x}} = e$$ Here, let \(x = \frac{1}{2n}\). As \(n\) approaches infinity, \(x\) approaches \(0\). Hence, using the property, we get: $$\lim_{n \to \infty} \left(1+\frac{1}{2n}\right)^{\frac{n}{2}} = \lim_{x \to 0}(1+x)^{\frac{1}{2x}}$$ Applying the limit: $$\lim_{x \to 0}(1+x)^{\frac{1}{2x}} = e^{\frac{1}{2}}$$
05

Conclusion

The limit of the given sequence as n approaches infinity is: $$\lim_{n \to \infty} \sqrt{\left(1+\frac{1}{2n}\right)^{n}} = e^{\frac{1}{2}}$$

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Key Concepts

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

Convergence and Divergence
Understanding the concepts of convergence and divergence is essential when studying sequences in mathematics. A sequence is said to converge if it approaches a specific limit as we go to infinity. In simpler terms, even though the sequence might keep on going forever, it settles down towards a single value that it gets closer and closer to.

On the other hand, a sequence diverges when it does not approach any particular value but instead keeps changing without settling down. This includes sequences that increase or decrease without bound, oscillate between values, or do not show any clear pattern at all.

In our exercise, we worked through the problem to determine whether the given sequence converges or diverges as it tends to infinity. The application of a limit to the sequence was the key step in establishing its behavior. Through calculation, we found that the sequence does indeed converge to a specific value, indicating that the sequence is not diverging.
Exponential Functions
An exponential function is a mathematical expression in which a variable appears in the exponent. A typical form of an exponential function is \( f(x) = a^{x} \) , where \( a \) is a positive constant known as the base of the exponential function.

Exponential functions are widely used across various fields of science and mathematics because of their unique property of rapid growth or decay. In our problem, we saw how the sequence had an exponential form since the variable n appeared in the exponent. By understanding the behavior and limits of exponential functions, we could correctly apply a known mathematical property to find the limit of our sequence, which is crucial when approaching problems that involve calculating the growth or decay of quantities over time.
Limits and Continuity
Limits are a fundamental concept in calculus, and they help define and understand the behavior of functions and sequences as they approach specific points or infinity. The limit expresses the value that a function or sequence 'aims for' as the input approaches some point.

Continuity, on the other hand, means that a function does not make any sudden jumps or breaks; it flows smoothly. If a function is continuous at a particular point, then its limit at that point is simply the function's value there.

In our exercise, by re-writing the sequence in a form that allowed us to apply the concept of limits, we were able to determine the behavior as n becomes very large. The notion of continuity implies that if we zoom in on any part of a continuous function, it will look like an unbroken line. This concept ensures that when we deal with the limit of a continuous function, we can approach it with the assurance that it will behave predictably, without any unexpected jumps or breaks.
Infinite Sequences
An infinite sequence is an ordered list of numbers that continues indefinitely. While the individual elements in an infinite sequence may vary, we're often interested in the overall trend or pattern as we move towards the later terms of the sequence.

In the context of our exercise, we handled an infinite sequence defined by a particular formula involving n, which represents the position in the sequence. By looking at the behavior of the sequence as n becomes infinitely large, we learn about the sequence's long-term trend. This allows us to understand sequences that model real-world processes, like population growth, finance, or physics, which often rely on predicting long-term behavior based on the properties of infinite sequences.

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Most popular questions from this chapter

Consider a wedding cake of infinite height, each layer of which is a right circular cylinder of height 1. The bottom layer of the cake has a radius of \(1,\) the second layer has a radius of \(1 / 2,\) the third layer has a radius of \(1 / 3,\) and the \(n\) th layer has a radius of \(1 / n\) (see figure). a. To determine how much frosting is needed to cover the cake, find the area of the lateral (vertical) sides of the wedding cake. What is the area of the horizontal surfaces of the cake? b. Determine the volume of the cake. (Hint: Use the result of Exercise 66.) c. Comment on your answers to parts (a) and (b).

Consider the geometric series \(S=\sum_{k=0}^{\infty} r^{k}\) which has the value \(1 /(1-r)\) provided \(|r|<1\). Let \(S_{n}=\sum_{k=0}^{n-1} r^{k}=\frac{1-r^{n}}{1-r}\) be the sum of the first \(n\) terms. The magnitude of the remainder \(R_{n}\) is the error in approximating \(S\) by \(S_{n} .\) Show that $$ R_{n}=S-S_{n}=\frac{r^{n}}{1-r} $$

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the limit of the sequence or state that the sequence diverges. $$a_{n+1}=4 a_{n}\left(1-a_{n}\right) ; a_{0}=0.5$$

A fishery manager knows that her fish population naturally increases at a rate of \(1.5 \%\) per month, while 80 fish are harvested each month. Let \(F_{n}\) be the fish population after the \(n\) th month, where \(F_{0}=4000\) fish. a. Write out the first five terms of the sequence \(\left\\{F_{n}\right\\}\). b. Find a recurrence relation that generates the sequence \(\left\\{F_{n}\right\\}\). c. Does the fish population decrease or increase in the long run? d. Determine whether the fish population decreases or increases in the long run if the initial population is 5500 fish. e. Determine the initial fish population \(F_{0}\) below which the population decreases.

The Greeks solved several calculus problems almost 2000 years before the discovery of calculus. One example is Archimedes' calculation of the area of the region \(R\) bounded by a segment of a parabola, which he did using the "method of exhaustion." As shown in the figure, the idea was to fill \(R\) with an infinite sequence of triangles. Archimedes began with an isosceles triangle inscribed in the parabola, with area \(A_{1}\), and proceeded in stages, with the number of new triangles doubling at each stage. He was able to show (the key to the solution) that at each stage, the area of a new triangle is \(\frac{1}{8}\) of the area of a triangle at the previous stage; for example, \(A_{2}=\frac{1}{8} A_{1},\) and so forth. Show, as Archimedes did, that the area of \(R\) is \(\frac{4}{3}\) times the area of \(A_{1}\).

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