Determine $\displaystyle{\lim_{n\to\infty} x_n}$ if $$\left(1+\frac{1}{n}\right)^{n+x_n}=e,\forall n\in \mathbb{N} $$

I have typed 2 methods giving two different answers

Method 1

$$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n+x_n}=\lim_{n\to\infty}e\\\implies \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}\cdot\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{x_n}=e\\\implies e\cdot\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{x_n}=e \\\implies \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{x_n}=1$$

We know: $$\lim_{x\to a}\left(f\left(x\right)^{g\left(x\right)}\right)=L\Rightarrow\ln L=g\left(x\right)\lim_{x\to a}\ln\left(f\left(x\right)\right)$$ $$\therefore 0=x_n\lim_{n\to\infty}(\ln(1+\frac{1}{n}))$$ Now $\lim_{n\to\infty}(\ln(1+\frac{1}{n}))=0$ therefore $x_n$ can be anything.

Method 2

Take log both sides and get $$\left(x_n+n\right)\ln\left(1+\frac{1}{n}\right)=1$$ Also as $$t=\frac{1}{n};n\to\infty;t\to0$$ Now $$x_n=\frac{1}{\ln\left(1+t\right)}-\frac{1}{t}=\frac{t-\ln\left(1+t\right)}{t^2\left(\frac{\ln\left(1+t\right)}{t}\right)}$$ $$\lim_{n\to\infty}x_n=\lim_{t\to0}\frac{t-\ln\left(1+t\right)}{t^2\left(\frac{\ln\left(1+t\right)}{t}\right)}=\lim_{t\to0}\frac{t-\ln\left(1+t\right)}{t^2}\cdot\lim_{t\to0}\frac{1}{\frac{\ln\left(1+t\right)}{t}}=\frac{1}{2}\cdot1$$

(Used certain standard limits which you can solve by series or wikipedia for more methods.)

Please help.

(By the way I typed the note in Mathquill. Really nice to use)

  • 1
    $\begingroup$ For any real $\;c\;$ , it can be $\;x_n=c\;\;\forall n\;$ , and thus $\;\lim x_n\;$ indeed cannot be determined. $\endgroup$ – DonAntonio Jun 1 '14 at 12:32
  • $\begingroup$ @DonAntonio See Method 2 which gives $x_\infty = \frac{1}{2}$ $\endgroup$ – Megh Parikh Jun 1 '14 at 12:34
  • 2
    $\begingroup$ In the first case, you are only using that $\lim(1+1/n)^{n+x_n}=e$, so you can't determine the limit from that information. You have more information than that. For example, if $x_n=\sqrt{n}$, then $\lim(1+1/n)^{n+x_n}=e$. If you forget some of the information of your problems, then you won't always be able to solve your problem.... $\endgroup$ – Thomas Andrews Jun 1 '14 at 12:55
  • 1
    $\begingroup$ The point is, you are given a very specific sequence - as you know, the $x_n$ are actually determined. In the first approach, you are only using a property of this sequence, which does not fully determine the sequence. There are plenty of other sequences that have this property, so what that means is that the property is not enough to conclude the limit you are looking for. $\endgroup$ – Thomas Andrews Jun 1 '14 at 13:01
  • 1
    $\begingroup$ You are still just saying that if we only know $\lim_{n\to\infty}(!+1/n)^{n+x_n}=e$, we can't determine $\lim x_n$ for this information. Which is true. That doesn't meant tha limit doesn't exist. There is some really sloppy notation - e.g., $(\to 1) ^ {finite} = 1$ is not math notation, so it is hard to tell what you mean here... $\endgroup$ – Thomas Andrews Jun 1 '14 at 13:22

You are still solving two different problems. In the first case, you are saying:

If I know that $$\lim_{n\to\infty} \left(1+\frac1n\right)^{n+x_n} = e$$ can I determine $\lim_{n\to\infty} x_n$?

Your conclusion is correct, you cannot determine $\lim x_n$ based in this information. For example, if $\forall n x_n=0$, then $\lim_{n\to\infty} \left(1+\frac1n\right)^{n+x_n} = e$, and $\lim_{n\to\infty}x_n = 0$, while if $\forall n x_n=1$, it still satisfies the condition.

But the second approach says:

If I know that $$\left(1+\frac1n\right)^{n+x_n} = e$$ can I determine $\lim_{n\to\infty} x_n$?

This is a lot more information. This entirely specifies the values $x_n$, while the previous condition only gives you some really loose information about the sequence. So it is no surprise that you can conclude more about $x_n$ if you have the entire sequence rather than just a property that can match lots of different sequences.

  • $\begingroup$ Thats precisely what my method 2 is. I am asking what is wrong with method 1? $\endgroup$ – Megh Parikh Jun 1 '14 at 12:52
  • $\begingroup$ Ah yes, jumped the gun. $\endgroup$ – Thomas Andrews Jun 1 '14 at 12:53
  • $\begingroup$ Edited answer to answer the actual question asked. $\endgroup$ – Thomas Andrews Jun 1 '14 at 13:39
  • $\begingroup$ Method 1 is nonsense because of two reasons: 1) the last step is vague and makes no sense (the part about "finite") and 2) the conclusion is invalid. It's like saying you are trying to solve $x^2=1$ and notice that if $x^2=1$ then $x=x$ and then you conclude that since $x=x$ for every $x$ that $x$ can be anything. Sure, $x=x$ for every $x$, but if $x=x$ it doesn't mean $x^2=1$. The first equation implies the other, but they are not equivalent. $\endgroup$ – Seth Jun 1 '14 at 14:18
  • $\begingroup$ Along with Seth's answer I finally understood what you were trying to say. $\endgroup$ – Megh Parikh Jun 2 '14 at 9:49

We have $$ \left(1+\frac{1}{n}\right)^{x_n} =e\left(1+\frac{1}{n}\right)^{-n}$$ hence $$x_n =\frac{1}{\ln\left(1+\frac{1}{n}\right)} -n$$ and therefore $$\lim_{n\to\infty} x_n =\frac{1}{2}.$$

  • $\begingroup$ How did you reach from 2nd to 3rd step? Is it a standard limit I am unaware of? $\endgroup$ – Megh Parikh Jun 1 '14 at 13:20
  • $\begingroup$ @MeghParikh: Asymptotic expansion $\endgroup$ – user21820 Jun 1 '14 at 13:21
  • $\begingroup$ You can use L'Hospital rule. $\endgroup$ – user110661 Jun 1 '14 at 13:22
  • $\begingroup$ @FisiaiLusia: Yes that would suffice here too, but in general asymptotic expansion can do everything and provides bounds on the error terms to any desired order, besides being intuitive. $\endgroup$ – user21820 Jun 1 '14 at 13:24
  • $\begingroup$ Yes this is true but can you pinpoint what is wrong with Method 1 in my question(sorry but I donot know Asymptotic expansion and used L'Hospital rule) $\endgroup$ – Megh Parikh Jun 1 '14 at 13:25


In your first method, you wrote $\lim_{n\to\infty}(1+\frac{1}{n})^{x_n}$ without proving its existence, so that line is already incorrect, even though it turns out to be still true. In a later line you wrote an expression where $x$ occurred outside a limit that binds $x$. That is just nonsense because you cannot talk about the value of a variable outside its scope.

In your second method, you got the correct answer, but the method is still not quite right. You did not prove that you can expand the domain on which you take limits to a punctured neighbourhood of $0$ (small non-zero reals). Originally the limit was for integer $n$, which would correspond to reciprocal values of $t$, which certainly do not cover all small non-zero reals. So if you obtain a limit of your transformed expression, like here, the answer would be correct because you did cover all reciprocals, but if you do not obtain a limit, it does not mean that the original sequence also does not have a limit.

Asymptotic expansion

As $n \to \infty$:

  $x_n = \dfrac{\ln(e)}{\ln(1+\frac{1}{n})} - n \in \dfrac{1}{\frac{1}{n}-\frac{1}{2n^2}+Θ(\frac{1}{n^3})}-n$ because $\frac{1}{n} \in o(1)$

  $ = \dfrac{n}{1-(\frac{1}{2n}+Θ(\frac{1}{n^2}))} ⊆ n\bigg(1+(\frac{1}{2n}+Θ(\frac{1}{n^2}))+Θ\left((\frac{1}{2n}+Θ(\frac{1}{n^2}))^2\right)\bigg)-n$ because:

    $\frac{1}{2n}+Θ(\frac{1}{n^2}) ⊆ o(1)$

  $ ⊆ n(1+\frac{1}{2n}+Θ(\frac{1}{n^2}))-n = \frac{1}{2}+Θ(\frac{1}{n})$

I encourage you to learn asymptotic expansion because it works on many limits with no ingenuity or tricks necessary.

  • $\begingroup$ That first line in his answer is a pretty elementary conclusion. If $\lim a_nb_n=ab$ and $\lim a_n=a\neq 0$ then $\lim b_n=b$. $\endgroup$ – Thomas Andrews Jun 1 '14 at 14:37
  • $\begingroup$ @ThomasAndrews: I know, but the way he wrote it suggests that he didn't know the exact conditions, and it is very easy for people to make mistakes if they don't always check that the expression they write down actually exists. $\endgroup$ – user21820 Jun 1 '14 at 14:43

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