# Solve Limits of Sequences: e2 when n=∞

• Fernando Rios
In summary, the book says that the answer to the question of what the limit of a sequence is, is e2, but this answer is not always easy to find. It can be found using the logarithm, but this requires taking the logarithm of a function which is continuous.
Fernando Rios
Homework Statement
In the following problems, find the limit of the given sequence as n→∞.
Relevant Equations
(1 + n^2)^(1/ (ln n))
If n is ∞, then ln (n) = ln (∞) = ∞

Then, 1/∞ = 0

Any number raised to "0" = 1, so the answer should be 1. However the book says the answer is e2. Could you provide me some help?

If we remember the definition of Euler's number, we have:
$$e \equiv \lim_{n \rightarrow \infty}\left(1 +\frac{1}{n}\right)^n$$

So, as we can see, this is a sequence, which has members that are all bigger than one, yet they are tending towards ##1^\infty## type of limit. This limit is generally undefined, in a sense that you have to inspect it more clearly. For example, if we have a sequence which has a limit ##1^\infty##, then logarithm of that sequence would have limit ##0 \times \infty##. Your limit is ##\infty^0## which would also turn into a ##0 \times \infty## type of limit if you take logarithm of the sequence. Now you might say, this should be zero, right? But the deal with the limits is that it's not exactly zero, it is something that is tending towards zero, multiplied by something that is increasing without bound. So the question of what the product should be, depends on how fast this one term is converging to zero, and how fast the other term is diverging. Imagine you have something that diverges exponentially, multiplied by something that linearly converges to zero, for example a sequence defined by ##a_n = 2^n \times \frac{1}{n}##. If you try writing it term by term, you will see that it diverges. But then look at ##a_n = n \times e^{-n}##. This one tends to zero. And both of those are of form ##0 \times \infty##.

So the point of the story is that ##1^\infty## is not equal to ##1##(same goes for ##\infty^0##), it must be investigated further. In fact, as we can see, we defined Euler's number ##e## to be a limit of sequence of that form. Which gives you the hint about your exercise. For example, you probably find a limit ##0 \times \infty## easier to deal with than this limit, so maybe there is a way to take the logarithm of this? Hope that helps.

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Delta2
Fernando Rios said:
Homework Statement: In the following problems, find the limit of the given sequence as n→∞.
Homework Equations: (1 + n^2)^(1/ (ln n))

If n is ∞, then ln (n) = ln (∞) = ∞
No, ##\infty## is not amongst the real numbers, so you can't do arithmetic with it, or use it in algebraic expressions.
Fernando Rios said:
Then, 1/∞ = 0
Also no, for the reason given above.
Fernando Rios said:
Any number raised to "0" = 1
Any finite number raised to the power 0 is 1, but ##1 + n^2## is not finite in the limit you're working with.
Fernando Rios said:
, so the answer should be 1. However the book says the answer is e2. Could you provide me some help?
Your textbook should have several examples of working with a limit of this type. These limits usually involve taking the log to get an expression in the form that you can use L'Hoptal's Rule on.
Antarres said:
This limit is generally undefined
Right, but a limit of the sort ##[1^\infty]## is called an indeterminate form, as are ##[\infty - \infty]##, ##[\frac \infty \infty]##, and several others.

Delta2
Mark44 said:
Right, but a limit of the sort ##[1^\infty]## is called an indeterminate form, as are ##[\infty - \infty]##, ##[\frac \infty \infty]##, and several others.
Yeah that's what I meant, just didn't use the proper English term for it.

As others have pointed out your limit is of the form ##{+\infty}^{0}## which is indeterminate. To work with this limit take the logarithm of it and then use L'Hospital rule.

You ll find that the limit of the logarithm is equal to $$\lim_{n \to \infty}\frac{2n^2}{1+n^2}$$ which is equal to 2. Then because the logarithmic function is continuous you can use the property that ##\lim\ln f(n)=\ln \lim f(n)## where ##f(n)## your original function. Thus you can conclude that the limit of the original function is ##e^2##.

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I think the identity ##a^b = e^{bln a} ## should help in here.

Antarres
Thank you all for your responses.

WWGD

## What is the formula for solving limits of sequences when n approaches infinity?

The formula for solving limits of sequences when n approaches infinity is lim(n→∞) an = L, where an represents the sequence and L represents the limit as n approaches infinity.

## What is the concept of infinity in mathematics?

In mathematics, infinity refers to a quantity or value that is larger than any number that can be represented. It is often used to describe the concept of something without limits or boundaries.

## What is the significance of the number e in solving limits of sequences?

The number e, also known as Euler's number, is an important constant in mathematics that is used to represent exponential growth. It plays a significant role in solving limits of sequences as it is often the limit of many exponential sequences.

## How do you determine the limit of a sequence as n approaches infinity?

To determine the limit of a sequence as n approaches infinity, you can evaluate the expression by substituting infinity for n and simplifying the resulting expression. If the resulting expression approaches a finite value, then that value is the limit of the sequence.

## Can limits of sequences be solved using other methods besides substituting n with infinity?

Yes, there are various other methods for solving limits of sequences, such as the squeeze theorem, the ratio test, and the root test. These methods can be used to determine the limit of a sequence without substituting infinity for n.

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