- #1

- 7

- 0

i am happy to be one of ur friends

just we have challenge because of this limit

so i hope that someone donate to solve it

thanx

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- Thread starter kahlan
- Start date

- #1

- 7

- 0

i am happy to be one of ur friends

just we have challenge because of this limit

so i hope that someone donate to solve it

thanx

- #2

- 22,129

- 3,298

I would suggest changing this in a 0/0 or an [tex]\infty / \infty[/tex] and applying L' Hopital.

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- #3

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But it looks obvious that the limit does not exist !!!

- #4

- 22,129

- 3,298

Ow yes, you're right. It obviously doesnt exist. I read one of the parantheses wrong.

- #5

HallsofIvy

Science Advisor

Homework Helper

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[tex]\lim_{x\to\infty}x[(1+ 1/x)^x- e][/tex]

It is well known that [itex](1+ 1/x)^x[/itex] goes to e so that (1+ 1/x)^x- e would go to 0- the additional x outside the braces would give an indeterminant form of "infinity* 0".

However, you have the "x" inside the braces and -e outside. [itex]\lim_{x\to\infty}x(1+ 1/x)^x[/itex] is of the form "infinity*e" which does not converge.

- #6

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- 0

[tex]\lim_{x\to\infty}x[(1+ 1/x)^x- e][/tex]

- #7

- 7

- 0

it sould be like [tex]\lim_{x\to\infty}x[(1+ 1/x)^x- e][/tex]

sorry guys

as well as answer will be 1

sorry guys

as well as answer will be 1

- #8

- 7

- 0

i am happy to be one of ur friends

just we have challenge because of this limit

so i hope that someone donate to solve it

thanx

sorry

- #9

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- 0

You have:

x[(1 + 1/x)^x] - e = x[exp(x*ln(1+1/x))-e)

Then you use h=1/x, what gives:

1/h*[exp(ln(1+h)/h)-e)

=1/h*[exp(1/h*(h-h²/2+h^3/3+o(h))-e)]

=1/h*[exp(1-h/2+h²/3+o(h))-e)]

=1/h*[e(exp(-h/2+h²/3+o(h))-1)]

=1/h*[e(1-h/2+h²/3+o(h)-1)

=1/h*[e(-h/2+h²/3+o(h)]

=e/h*(-h/2+h²/3+o(h)]

=-e/2+h/3+o(h)

As x→∞, then h→0

Finally, you get:

lim x→∞ (x[(1 + 1/x)^x] - e) = lim h→0 (-e/2+h/3+o(h)) = -e/2

- #10

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[tex]x\bigl((1 + 1/x)^x - e\bigr) = -\frac e 2 + \frac{11 e}{24}\frac 1 x - O\bigl(1/x^2\bigr)[/tex]

or

[tex]\bigl((1 + h)^{1/h} - e\bigr)/h = -\frac e 2 + \frac{11 e}{24}h - O\bigl(h^2\bigr)[/tex]

In the limit x→∞ or h→0, these become -e/2.

Somehow you did obtain the right limit.

- #11

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[tex]x\bigl((1 + 1/x)^x - e\bigr) = -\frac e 2 + \frac{11 e}{24}\frac 1 x - O\bigl(1/x^2\bigr)[/tex]

or

[tex]\bigl((1 + h)^{1/h} - e\bigr)/h = -\frac e 2 + \frac{11 e}{24}h - O\bigl(h^2\bigr)[/tex]

In the limit x→∞ or h→0, these become -e/2.

Somehow you did obtain the right limit.

You're right, thanks for noticing, I've forgotten something in the series expansion of exp(u), actually the u²/2 part with u=-h/2+h²/3+o(h²), which actually gives the element h²/8:

1/h*[exp(ln(1+h)/h)-e)

=1/h*[exp(1/h*(h-h²/2+h^3/3+o(h²))-e)]

=1/h*[exp(1-h/2+h²/3+o(h²))-e)]

=1/h*[e(exp(-h/2+h²/3+o(h²))-1)]

=1/h*[e(1-h/2+h²/3+o(h²)-1)

=1/h*[e(-h/2+h²/3+o(h²)]

=e/h*(-h/2+11h²/24+o(h²)

=-e/2+11h/24+o(h)

And then you find the limit -e/2.

PS: Where can I find a tutorial/help section on this forum to know how to write series expansion formulas like what you've done?

- #12

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This thread: [thread=546968]LaTeX Guide: Include mathematical symbols and equations in a post[/thread].PS: Where can I find a tutorial/help section on this forum to know how to write series expansion formulas like what you've done?

One last comment about this thread:

Look at the dates when posting. This thread was last active over a year ago. In general it isn't a good idea to raise old threads from the dead. In this case it was OK because (a) there was no correct resolution, and (b) a year is old but not terribly old. On the other hand, if you come across a seven year old thread with no resolution, don't get the seven year itch to resurrect it. Just let it lie.

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