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?

If you mean the limit as $n\to ∞$, then the limit is equal to $\frac{1}{e}$ — KR136, Apr 16 '16 at 23:59
Flip it over, look at the limit of $(n+1)/n)^n$, that is, $(1+1/n)^n$. Familiar? — André Nicolas, Apr 17 '16 at 00:01
The problem with you logic is that both the base AND the exponent depend on $n$ simultaneously. — John Joy, Apr 17 '16 at 06:29
I have changed $\lim\limits_{x\to {\infty}} (\frac{n}{1+n})^n$ in your post to $\lim\limits_{n\to {\infty}} (\frac{n}{1+n})^n$. (I suppose $x$ instead of $n$ was a typo. And also this is what you wrote in the title, so this is probably what you wanted to ask.) — Martin Sleziak, Apr 17 '16 at 10:17

score 6 · Answer 1 · answered Apr 16 '16 at 23:59

6

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.

answered Apr 16 '16 at 23:59

Mnifldz

12,880

Didn't think of that. Thanks. – CoolKid Apr 17 '16 at 00:01

score 5 · Answer 2 · answered Apr 17 '16 at 00:01

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 $1/e$.

Bernard · Answer 3 · 2016-04-17T09:33:22.570

1

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 $\;\dfrac1{\mathrm{e}}$.

edited Apr 17 '16 at 09:33

answered Apr 17 '16 at 00:05

Bernard

175,478

It is not really useful not use notation such as $\sim_\infty$ when the OP is learning how to deal with indeterminate forms. – Pedro Apr 17 '16 at 06:48
@Pedro Tamaroff: This depends on the O.P.'s syllabus. In my country, this is taught to first year students. My opinion is it's never too soon to learn efficient methods in simple cases. – Bernard Apr 17 '16 at 09:39

score 1 · Answer 4 · answered Apr 17 '16 at 06:45

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 $$\lim_{n\to\infty}\bigg(1-\frac{1}{n+1}\bigg)^n = \lim_{h\to 0}\bigg(1+h\bigg)^{-\frac{1}{h}-1}=\frac{1}{\lim_{h\to 0}(1+h)^{1/h}}\cdot\frac{1}{\lim_{h\to 0}(1+h)^1}$$ $$=\frac{1}{e}\cdot\frac{1}{1+0}=\frac{1}{e}$$

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

4 Answers4