Solve Limits of Sequences: e2 when n=∞

[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?

 

[Antarres]

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.

 

[Mark44]

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.

Then, 1/∞ = 0

Also no, for the reason given above.

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.

, 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.

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.

 

[Antarres]

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.

 

[Delta2]

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 $e^2$.

 

[WWGD]

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

 

[Fernando Rios]

Thank you all for your responses.