Lim x→∞ (x[(1 + 1/x)^x] - e)

  • Thread starter kahlan
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  • #1
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how are u every one there
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
 

Answers and Replies

  • #2
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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
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Ow yes, you're right. It obviously doesnt exist. I read one of the parantheses wrong.
 
  • #5
HallsofIvy
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Kahlan, the "wrong parentheses" micromass was seeing was
[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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[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
 
  • #8
7
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how are u every one there
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
2
0
The limit does exist.

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
D H
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scichem, you're being a bit sloppy with braces and parentheses, so it's hard to see where you went wrong. You should have obtained

[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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scichem, you're being a bit sloppy with braces and parentheses, so it's hard to see where you went wrong. You should have obtained

[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+h²/3+h²/8+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
D H
Staff Emeritus
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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?
This thread: [thread=546968]LaTeX Guide: Include mathematical symbols and equations in a post[/thread].


Edit
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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