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# Doubt in calculus limit problem

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.)

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

Note by Megh Parikh
3 years, 1 month ago

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I 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 $$\left( 1 + \frac{1}{n} \right) ^{Y_n} =1$$. Staff · 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? · 3 years, 1 month ago

The simple english version is that your implications only work one way (as indicated by your forward arrows). Hence, the given condition implies your final conclusion, but not necessarily the other way around.

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. Staff · 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? · 3 years, 1 month ago

Simply put, if you do not prove that your implications work both ways, then it need not be true that the implications work both ways.

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

Yes I understood thanks · 3 years, 1 month ago

It's true that $$x_n$$ can be any constant, there is no problem with that. · 3 years, 1 month ago

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

Calvin says that it can be any other finite constant, so it is :D. · 3 years, 1 month ago

I did not say that "the limit of $$x_n$$ (as defined in the question) is any finite constant".

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. Staff · 3 years, 1 month ago

Oh...but then any fixed constant will work, won't it? · 3 years, 1 month ago

No. $$x_n$$ is defined in a very unique way in the question.

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

Ah, I checked the problem and now I understand. · 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 $$\lim_{n\to\infty} x_n =anything$$ · 3 years, 1 month ago

Very very false. You will get issues if these sequences approach 0 or $$\infty$$. Look through the proofs, and see why these edges cases are important.

See for example They told me that $$0^0 = 1$$, but why? . Staff · 3 years, 1 month ago

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