- #1

- 368

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I can't move with this limit:

[tex]

\lim_{n \rightarrow \infty} n \left( \left( 1 + \frac{1}{n} \right)^{n} - e \right)

[/tex]

Could someone help me please? Or some hint...

But no l'Hospital please.

Thank you

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- Thread starter twoflower
- Start date

- #1

- 368

- 0

I can't move with this limit:

[tex]

\lim_{n \rightarrow \infty} n \left( \left( 1 + \frac{1}{n} \right)^{n} - e \right)

[/tex]

Could someone help me please? Or some hint...

But no l'Hospital please.

Thank you

- #2

- 1,357

- 0

[tex]

\lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^{n}

[/tex]

is. It should all be clear from then on.

- #3

- 368

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[tex]

\lim_{n \rightarrow \infty} n \left( e - e \right) = \lim_{n \rightarrow \infty} n \left ( 0 \right) = 0.\infty

[/tex]

Which is undefined...

- #4

shmoe

Science Advisor

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If you are opposed to using series but have learned them, it's probably possible to modify the above into a squeeze limit type of proof. If you haven't learned Taylor series yet, I'll have to think of a more elementary method, though it will probably be a thinly disguised version of the above.

- #5

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- #6

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twoflower said:

[tex]

\lim_{n \rightarrow \infty} n \left( e - e \right) = \lim_{n \rightarrow \infty} n \left ( 0 \right) = 0.\infty

[/tex]

Which is undefined...

you cannot sub in the value for e in between the limit like that

- #7

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- 2

split the limit so you have the limits of n(1-1/n)^n and n e subtracted from each other.

now the limit of n (1-1/n)^n is infinity while the limit of n E is also infinity

however the first one converges to e more slowly that E already does

so n E >n(1-1/n)^n and the limit is negative infinity

- #8

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The limit should be -e/2...

- #9

- 1,444

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twoflower said:The limit should be -e/2...

check what you entered into maple/mathematica/matlab... it isn't -e/2, it is -infinity

- #10

shmoe

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stunner5000pt said:

split the limit so you have the limits of n(1-1/n)^n and n e subtracted from each other.

now the limit of n (1-1/n)^n is infinity while the limit of n E is also infinity

however the first one converges to e more slowly that E already does

so n E >n(1-1/n)^n and the limit is negative infinity

No, this won't work. You're essentially saying infinity-infinity=-infinity because the first infinity is getting there slower. This is bunk. An indeterminate infinity-infinity limit form can potentially equal anything we like(-e/2 is correct in this case, done by hand with the method I suggested).

twoflower-series is likely what's intended then. I can't see another way that isn't unnecessarily complicated.

- #11

NateTG

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twoflower said:Hi all,

I can't move with this limit:

[tex]

\lim_{n \rightarrow \infty} n \left( \left( 1 + \frac{1}{n} \right)^{n} - e \right)

[/tex]

Could someone help me please? Or some hint...

If you can show that

[tex]-\frac{e}{2n} - \epsilon_1(n) \leq \left( 1 + \frac{1}{n} \right)^{n} - e \leq -\frac{e}{2n} + \epsilon_2(n)[/tex]

Where [itex]\epsilon(n)[/itex] is some expression that goes to zero faster than

[itex]\frac{1}{n}[/itex] then you're set, so perhaps you should look at the convergence of [itex]\left( 1 + \frac{1}{n} \right)^{n} [/itex]

- #12

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stunner5000pt said:check what you entered into maple/mathematica/matlab... it isn't -e/2, it is -infinity

I tried it now. Maple gives -e/2

- #13

shmoe

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The obvious approach to me involves some bounds on log(1+x) and e^x that I would prove using series.

- #14

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shmoe said:

The obvious approach to me involves some bounds on log(1+x) and e^x that I would prove using series.

Ok, so you say guys that the most natural way to find this limit is to use Taylor series. If it's right, I will skip this one. Or does anyone have simplier approach to show?

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