What is $\lim\limits_{n\to {\infty}} (\frac{n}{1+n})^n$?

Question

What is $\lim\limits_{n\to {\infty}} (\frac{n}{1+n})^n$.

Is it possible to write the function $f(x)=x^n$ and since we know $\frac{n}{1+n}\to 1$, so $f(\frac{n}{1+n})\to 1^n=1$. So the limit it $1$. Is it correct. If so, is there an easier way to do it?

Answer 1

Observe that

$$ \left ( \frac{n}{n+1} \right )^n \;\; =\;\; \left ( 1 + \frac{1}{n} \right )^{-n} \;\; =\;\; \frac{1}{\left (1 + \frac{1}{n} \right )^n}. $$

Since $\lim_{n\to \infty} \left ( 1 + \frac{1}{n} \right )^n$ is a well known limit this should be pretty easy at this point.

Answer 2

If $f$ is a continuous function, one could say that $\displaystyle \lim_{n\to\infty} f\left( \frac n {1+n} \right) = f\left( \lim_{n\to\infty} \frac n {1+n} \right)$, but that's not what we have here: we have a different function $f$ for each value of $n$, so that as $n\to\infty$, the function $f$ keeps changing as $n$ grows.

You may recall that $\lim\limits_{n\to\infty} \left( 1 + \dfrac 1 n \right)^n = e$.

That's the same as $ \lim\limits_{n\to\infty} \left( \dfrac{n+1} n \right)^n = e$.

Here you have the limit of the reciprocal of that, and the reciprocal function is continuous, and that function does not keep changing as $n$ grows, so you end up with $1/e$.

Answer 3

No, it's not correct: your reasoning replaces a part of the expression with its limit, while letting the rest variable. This way you can prove the limit is any number.

To have the answer, it is enough to compute the limit of the log of the expression. Now $$\log\Bigl(\frac n{n+1}\Bigr)^n=n\log\Bigl(1-\frac 1{n+1}\Bigr)\sim_\infty n\cdot \frac1{n+1}\sim_\infty-1. $$ So the limit is $\;\dfrac1{\mathrm{e}}$.

Answer 4

I've always hated the formula $\lim_{n\to\infty}(1+\frac{1}{n})^n = e$, so I would always substitute $h=\frac{1}{n}$, giving me a new limit $\lim_{h\to 0}(1+h)^\frac{1}{h} = e$

In this problem we have $$\lim_{n\to\infty}\bigg(\frac{n}{n+1}\bigg)^n = \lim_{n\to\infty}\bigg(\frac{n+1-1}{n+1}\bigg)^n=\lim_{n\to\infty}\bigg(1-\frac{1}{n+1}\bigg)^n$$

Making the substitution $h = -\frac{1}{n+1}$ we have $$h=-\frac{1}{n+1}$$ $$n+1=-\frac{1}{h}$$ $$n=-\frac{1}{h}-1$$ Giving us $$\lim_{n\to\infty}\bigg(1-\frac{1}{n+1}\bigg)^n = \lim_{h\to 0}\bigg(1+h\bigg)^{-\frac{1}{h}-1}=\frac{1}{\lim_{h\to 0}(1+h)^{1/h}}\cdot\frac{1}{\lim_{h\to 0}(1+h)^1}$$ $$=\frac{1}{e}\cdot\frac{1}{1+0}=\frac{1}{e}$$