Waste less time on Facebook — follow Brilliant.
×

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.)


Please help.

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

Note by Megh Parikh
3 years, 6 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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 \( \left( 1 + \frac{1}{n} \right) ^{Y_n} =1 \).

Calvin Lin Staff - 3 years, 6 months ago

Log in to reply

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

Megh Parikh - 3 years, 6 months ago

Log in to reply

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.

Calvin Lin Staff - 3 years, 6 months ago

Log in to reply

@Calvin Lin Can you perhaps state how solution space is increased or share a note stating steps which we muust not take otherwise increased solution space?

Megh Parikh - 3 years, 6 months ago

Log in to reply

@Megh Parikh 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.

Calvin Lin Staff - 3 years, 6 months ago

Log in to reply

@Calvin Lin Yes I understood thanks

Megh Parikh - 3 years, 6 months ago

Log in to reply

It's true that \(x_n\) can be any constant, there is no problem with that.

Bogdan Simeonov - 3 years, 6 months ago

Log in to reply

@Bogdan Simeonov Now see the Method 2 in the note

Megh Parikh - 3 years, 6 months ago

Log in to reply

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.

Megh Parikh - 3 years, 6 months ago

Log in to reply

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

Bogdan Simeonov - 3 years, 6 months ago

Log in to reply

@Bogdan Simeonov 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.

Calvin Lin Staff - 3 years, 6 months ago

Log in to reply

@Calvin Lin Oh...but then any fixed constant will work, won't it?

Bogdan Simeonov - 3 years, 6 months ago

Log in to reply

@Bogdan Simeonov 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.

Calvin Lin Staff - 3 years, 6 months ago

Log in to reply

@Calvin Lin Ah, I checked the problem and now I understand.

Bogdan Simeonov - 3 years, 6 months ago

Log in to reply

@Calvin Lin 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 \(\lim_{n\to\infty} x_n =anything\)

Megh Parikh - 3 years, 6 months ago

Log in to reply

@Megh Parikh 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? .

Calvin Lin Staff - 3 years, 6 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...