What is the limit of this expression as n approaches infinity?

[aruwin]

Hello.
How do I find the limit of this term?

$$\lim_{{n}\to{\infty}}|\left(\frac{n}{n+1}\right)^{\!{n^2}}|^\frac{1}{n}$$

 

[Prove It]

Hello.
How do I find the limit of this term?

$$\lim_{{n}\to{\infty}}|\left(\frac{n}{n+1}\right)^{\!{n^2}}|^\frac{1}{n}$$

Well for starters, how can you simplify this expression?

 

[aruwin]

Well for starters, how can you simplify this expression?

This is how it is simplified (I skipped the lim symbol)$\left(\frac{n}{n+1}\right)^{\!{n}}$

 

[Prove It]

This is how it is simplified (I skipped the lim symbol)$\left(\frac{n}{n+1}\right)^{\!{n}}$

Yes. So now I would think about if it's possible to get it in the form $\displaystyle \begin{align*} \frac{0}{0} \end{align*}$ or $\displaystyle \begin{align*} \frac{\infty}{\infty} \end{align*}$ so that I can use L'Hospital's Rule.

To do this, write it as

$\displaystyle \begin{align*} \left( \frac{n}{n + 1} \right) ^n = \mathrm{e}^{ \ln{ \left[ \left( \frac{n}{n+1} \right) ^n \right] } } \end{align*}$

Now try to simplify it using logarithm laws.

 

[aruwin]

Yes. So now I would think about if it's possible to get it in the form $\displaystyle \begin{align*} \frac{0}{0} \end{align*}$ or $\displaystyle \begin{align*} \frac{\infty}{\infty} \end{align*}$ so that I can use L'Hospital's Rule.

To do this, write it as

$\displaystyle \begin{align*} \left( \frac{n}{n + 1} \right) ^n = \mathrm{e}^{ \ln{ \left[ \left( \frac{n}{n+1} \right) ^n \right] } } \end{align*}$

Now try to simplify it using logarithm laws.

What would $\frac{0}{0}and \frac{\infty}{\infty}$ mean?

 

[I like Serena]

What would $\frac{0}{0}and \frac{\infty}{\infty}$ mean?

They are meant to indicate that both numerator and denominator tend to $0$ respectively $\infty$. This is a precondition to apply l'Hopitals rule.

Instead of applying l'Hopitals rule, you can also simplify further to:
$$\lim_{n\to \infty} \frac{1}{\left(1+\frac{1}{n}\right)^{\!{n}}}$$

Now you can make use of the standard limit:
$$\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^{\!{n}} = e$$

 

[aruwin]

They are meant to indicate the both numerator and denominator tend to $0$ respectively $\infty$. This is a precondition to apply l'Hopitals rule.

Instead of applying l'Hopitals rule, you can also simplify further to:
$$\lim_{n\to \infty} \frac{1}{\left(1+\frac{1}{n}\right)^{\!{n}}}$$

Now you can make use of the standard limit:
$$\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^{\!{n}} = e$$

I haven't done l'hopitals rule in a while so I read about it on wiki.
But the simplification method is much easier. That means the limit is 1/e,
Thanks!

 

[Fantini]

You can also write $$\left( \frac{n}{n+1} \right)^n = \left( \frac{n+1-1}{n+1} \right)^n = \left( 1 - \frac{1}{n+1} \right)^n$$ and deduce the limit from here. :)