Determine \(\displaystyle{\lim_{n\to\infty} x_n}\) if \[\left(1+\frac{1}{n}\right)^{n+x_n}=e\]

I have typed 2 methods giving two different answers

## Method 1

\[\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n+x_n}=\lim_{n\to\infty}e\\\implies \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}\cdot\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{x_n}=e\\\implies e\cdot\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{x_n}=e \\\implies \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{x_n}=1\] Also as \(x_n\) was finite \((\to 1) ^ {finite} = 1\) and thus \(x_n\) cannot be determined.

## 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)

## Comments

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TopNewestI remember setting this question before - The Limit Of The Exponents.

The issue with your first method is that when you took limits in the first step, you increased the solution space of \(x_n\) from being an exact value, to being a much larger set. This is why you reach the conclusion that \(x_n\) cannot be determined. As you denoted, you only have forward implication signs, but not backward implication.

Note: It is true that if the limit of \(Y_n\) is \( \frac{1}{2} \) (or any other constant), then the limit of \( \left( 1 + \frac{1}{n} \right) ^{Y_n} =1 \). – Calvin Lin Staff · 3 years, 1 month ago

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– Megh Parikh · 3 years, 1 month ago

I don't understand what you want to say. Can you please explain in more simple English and less higher math jargon?Log in to reply

As an example, what you did was equivalent to saying that

\[ x = 1 \Rightarrow x ^2 = 1 \Rightarrow x^2 - 1 = 0 \Rightarrow (x-1)(x+1) + 0 \Rightarrow x = 1 \text{ or }-1. \]

and claiming that \( x = -1 \) is a solution to the original equation. This is not true, because the implications only work one way. We only have \( x = 1 \Rightarrow x = 1 \text{ or } -1 \) but we do not have \( x = 1 \text{ or } -1 \Rightarrow x = 1 \).

You can see how in the first implication, you increased the solution space from 1 element, to 2 elements. – Calvin Lin Staff · 3 years, 1 month ago

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– Megh Parikh · 3 years, 1 month ago

Can you perhaps state how solution space is increased or share a note stating steps which we muust not take otherwise increased solution space?Log in to reply

Whenever you manipulate equations (or inequalities), you need to be acutely aware of that. – Calvin Lin Staff · 3 years, 1 month ago

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– Megh Parikh · 3 years, 1 month ago

Yes I understood thanksLog in to reply

It's true that \(x_n\) can be any constant, there is no problem with that. – Bogdan Simeonov · 3 years, 1 month ago

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@Bogdan Simeonov Now see the Method 2 in the note – Megh Parikh · 3 years, 1 month ago

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– Megh Parikh · 3 years, 1 month ago

That's no true. Youu can use series expansion to get the answer as \(\frac{1}{2}\). Also I know another method to get this answer. Please wait for me to type it out today night.Log in to reply

– Bogdan Simeonov · 3 years, 1 month ago

Calvin says that it can be any other finite constant, so it is :D.Log in to reply

I said that IF the limit of \(Y_n\) (which is not the same as the sequence defined in the question) is any finite constant, then the limit of \( (1 + \frac{1}{n} ) ^{ Y_n} \) is 1. – Calvin Lin Staff · 3 years, 1 month ago

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– Bogdan Simeonov · 3 years, 1 month ago

Oh...but then any fixed constant will work, won't it?Log in to reply

It is likely that you are confused because I used the same variable throughout. Let me make an edit. – Calvin Lin Staff · 3 years, 1 month ago

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– Bogdan Simeonov · 3 years, 1 month ago

Ah, I checked the problem and now I understand.Log in to reply

– Megh Parikh · 3 years, 1 month ago

Well \(\displaystyle{\lim_{x\to a} f(x) ^{g(x)} = \lim{x\to a} f(x)^{\lim{x\to a} g(x)} }\) and so wont we get \(\lim_{n\to\infty} x_n =anything\)Log in to reply

See for example They told me that \(0^0 = 1 \), but why? . – Calvin Lin Staff · 3 years, 1 month ago

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