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Chapter 4: Problem 7

Does \(\lim _{n \rightarrow \infty}\left(\frac{1+1}{\sqrt{2}}\right)^{n}\) exist? Why?

Short Answer

Expert verified
The limit does not exist because \( \sqrt{2}^n \) grows unbounded as \( n \to \infty \).

Step by step solution

01

Identify the Expression

The expression we need to evaluate is \( \lim _{n \rightarrow \infty}\left(\frac{1+1}{\sqrt{2}}\right)^{n} \). Simplify the base of the exponent: \( \frac{1+1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \).
02

Analyze the Limit

We now have \( \lim _{n \rightarrow \infty} (\sqrt{2})^n \). Observe that since \( \sqrt{2} \approx 1.414 \), which is greater than 1, raising \( \sqrt{2} \) to increasing powers as \( n \) approaches infinity will cause the expression to grow indefinitely.
03

Conclusion on the Existence of Limit

Since \( (\sqrt{2})^n \) grows unbounded as \( n \to \infty \), the limit does not exist. For a limit to exist and be finite, the base of the exponential function should be less than or equal to 1.

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Key Concepts

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

Exponential Functions
Exponential functions are a fundamental type of mathematical function with the form \( a^n \), where \( a \) is a constant base, and \( n \) is the exponent. These functions are essential for describing growth and decay processes, such as compound interest or population growth.

When the base \( a \) is greater than 1, the exponential function represents exponential growth. This means that as \( n \) increases, the function's value grows rapidly and can become very large.
This behavior is distinct from other types of functions, like linear or quadratic functions, because exponential functions increase by a constant multiplier rather than a constant addition.
  • If the base \( a \) is between 0 and 1, the function models exponential decay.
  • Common bases for exponential functions include \( e \) (approximately 2.718), which is used in natural exponential functions, and 10, especially in logarithmic contexts.
Limit Existence
The existence of a limit is a critical concept in calculus and analysis. Determining whether a limit exists involves examining the behavior of a function as its input approaches a certain value, which could be a real number or infinity.

For a limit to exist:
  • The function must approach a particular finite value as the input approaches the target value.
  • The left and right limits, or the behavior from either side of the target, must coincide and equal the limit value.
If these conditions are not met, the limit either does not exist or may be infinite.

In exponential functions, if the base is greater than 1 and we evaluate the limit of its powers as the exponent goes to infinity, the function's value grows larger and larger without bound. Thus, the function does not converge to a particular finite value, indicating the limit does not exist in a finite sense.
Infinite Limits
Infinite limits occur when the values a function approaches become arbitrarily large, either positively or negatively, as the input approaches a certain value.

These limits describe scenarios where a function's behavior is unbounded, which doesn't satisfy the conditions for a finite limit.
  • If an exponential function has a base greater than 1, it tends to infinity as the input (or exponent) increases towards infinity.
  • This characteristic is why the exponential function \( (\sqrt{2})^n \) has an infinite limit when \( n \to \infty \).
  • An infinite limit essentially means that the function heads off to infinity, which is crucial when analyzing functions that grow rapidly.
Understanding infinite limits is important for interpreting function behavior at boundaries and for practical applications like assessing long-term growth trends.

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Most popular questions from this chapter

Prove that a sequence can have only one limit. Hint: Suppose that there is a sequence \(\left\\{z_{n}\right\\}\) such that \(z_{n} \rightarrow \zeta_{1}\) and \(z_{n} \rightarrow \zeta_{2}\). Show this implies \(\zeta_{1}=\zeta_{2}\) by proving that for all \(\varepsilon>0,\left|\zeta_{1}-\zeta_{2}\right|<\varepsilon\).

Prove that Newton's method always works for polynomials of degree 1 (functions of the form \(f(z)=a z+b\), where \(a \neq 0\) ). How many iterations are necessary before Newton's method produces the solution \(z=-\frac{1}{a}\) to \(f(z)=0 ?\)

Let \(z_{n}=r_{n} e^{i \theta_{n}} \neq 0\), where \(\theta_{n}=\operatorname{Arg}\left(z_{n}\right)\). (a) Suppose \(\lim _{n \rightarrow \infty} r_{n}=r_{0}\) and \(\lim _{n \rightarrow \infty} \theta_{n}=\theta_{0} .\) Show \(\lim _{n \rightarrow \infty} r_{n} e^{i \theta_{n}}=r_{0} e^{i \theta_{0}}\) (b) Find an example where \(\lim _{n \rightarrow \infty} z_{n}=z_{0}=r_{0} e^{i \theta_{0}}, \lim _{n \rightarrow \infty} r_{n}=r_{0}\), but \(\lim _{n \rightarrow \infty} \theta_{n}\) does not exist. (c) Is it possible to have \(\lim _{n \rightarrow \infty} z_{n}=z_{0}=r_{0} e^{i \theta_{0}}\), but \(\lim _{n \rightarrow \infty} r_{n}\) does not exist?

In the geometric series, show that if \(|z|>1\), then \(\lim _{n \rightarrow \infty}\left|S_{n}\right|=\infty\).

Let \(\sum_{k=1}^{\infty}\left(x_{n}+i y_{n}\right)=u+i v .\) If \(c=a+i b\) is a complex constant, show that \(\sum_{n=1}^{\infty}(a+i b)\left(x_{n}+i y_{n}\right)=(a+i b)(u+i v)\)
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