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I Is it possible to find the limit of (1+1/x)^x as x approaches -infinity?

  1. May 28, 2017 #1
    Is it possible to find the limit of (1+1/x)^x as x approaches minus infinity using only the fact that it is e if x approaches infinity?
     
  2. jcsd
  3. May 28, 2017 #2
  4. May 28, 2017 #3
    No, the answer is e. We have to find a reason without using LHospital's rule.
     
  5. May 28, 2017 #4

    fresh_42

    Staff: Mentor

    What happens, if you write ##y=-x## and consider ##y \to +\infty##?
     
  6. May 28, 2017 #5
    I thought about it but got stuck up. Then it becomes (1-1/y)-y as y approaches infinity which is not a standard formula.
     
  7. May 28, 2017 #6

    fresh_42

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    Maybe, but ##a^{-y}=(\frac{1}{a})^y## is.
     
  8. May 28, 2017 #7
    but 1+1/y is different from 1-1/y. That is where I am stuck.
     
  9. May 28, 2017 #8

    fresh_42

    Staff: Mentor

    For ##y \to \infty##? Isn't it only the difference between left and right?
     
  10. May 28, 2017 #9
    I am not sure if this helps but what if you multiply by the conjugate and then try to think why ##\lim_{y \to \infty} \left(1 - \frac{1}{y^2}\right)^y## should be 1.
     
  11. May 28, 2017 #10
    A great idea but I am not able to think of a proof that it is 1. If this is proved, then I agree that the original question is proved. But how to prove that? It raises other similar interesting questions like what about (1-1/y^3)^y, (1-1/y^4)^(y^2) etc.
     
  12. May 28, 2017 #11
    I think this problem can be solved in this way. Here is what I think:-

    let x = -1/y
    then, Lt (1+1/x)x where x→ -∞
    => Lt (1 - y)-1/y where y→ 0
    Now let -y = 1/p
    then, Lt (1 - y)-1/y where y→ 0
    => Lt (1+1/p)p where p→∞
    And this is equal to e.
     
  13. May 29, 2017 #12

    ehild

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    You have
    [tex]A=\lim _{x\rightarrow\infty} (1+1/x)^x=e[/tex]
    [tex]B=\lim _{x\rightarrow\infty} \left(\frac{1}{1-1/x}\right)^x[/tex]
    [tex]AB=\lim _{x\rightarrow\infty}\left(\frac{1+1/x}{1-1/x}\right)^x=\lim _{x\rightarrow\infty}(1+2/x)^x=\lim _{x\rightarrow\infty}\left((1+2/x)^{x/2}\right)^2=e^2[/tex]
     
  14. May 29, 2017 #13
    I am not able to figure out how 1+2/x came. Is it an approx?
     
  15. May 29, 2017 #14

    ehild

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    Yes, it is an approximation. 1/(1-a) ≈ 1+a and (1+a)2≈1+2a for |a|<<1 . As x-->∞, 1/x -->0 and you can substitute 1/(1-1/x) with (1+1/x). The second approximation is not needed as you have (1+1/x)x already.
    1/(1-a) is the sum of the geometric series 1+a+a2+.... . If a <<1, the higher order terms can be neglected.
     
  16. May 29, 2017 #15
    @ ehild, Great explanation. Thanks to all the respondents who took time to clarify things to me. I would like to know how to write math equations here, is there a link? Is there a way I can accept the answer and close as the raiser of this question?
     
  17. May 29, 2017 #16

    ehild

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    To write nice equations use LaTeX : https://www.physicsforums.com/help/latexhelp/
    There are useful symbols also under the ∑ key, and you can make subscript and superscript with the x2 and x2 keys.
    I think, closing a thread is possible in the Homework Forum only.
     
  18. May 29, 2017 #17

    jack action

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    This question has been intriguing me, so I've gave it a shot:
    $$\left(1+\frac{1}{x}\right)^x = \left(\frac{1+x}{x}\right)^x$$
    With ##\infty##:
    $$\left(\frac{1+\infty}{\infty}\right)^{\infty}$$
    With ##-\infty##:
    $$\left(\frac{1-\infty}{-\infty}\right)^{-\infty} = \left(\frac{\infty - 1}{\infty}\right)^{-\infty} = \left(\frac{\infty}{\infty - 1}\right)^{\infty}$$
    And one can see that:
    $$\frac{\infty}{\infty - 1} = \frac{1+\infty}{\infty}$$
    Where the numerator is just 1 + the denominator (like ##\frac{101}{100}## or ##\frac{1000}{999}##). Thus it should give the same answer when at infinity.
     
  19. May 29, 2017 #18

    fresh_42

    Staff: Mentor

    This shouldn't get on my desk for correction.
     
  20. May 29, 2017 #19

    Charles Link

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    @fresh_42 I agree. They could do a better job of solving this. Suggestion in computing ## y=## limit ## (1+\frac{1}{x})^x ## in a couple of the above posts is to take the natural log of this expression, (so that you have ## ln(y)=x \, ln(1+\frac{1}{x}) ##), use a Taylor expansion of ## ln(1+\frac{1}{x}) ##, (in the form of ## ln(1+u ) ## where ## u ## is small), and then take ## e^{ln(y )}=y ##.
     
    Last edited: May 29, 2017
  21. May 30, 2017 #20

    pwsnafu

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    Here's a direct method
    ##\lim_{x\to-\infty}\left(1+\frac{1}{x}\right)^x = \lim_{y\to\infty}\left(1-\frac{1}{y}\right)^{-y} = \lim_{y\to\infty}\frac{1}{\left(1-\frac{1}{y}\right)^y}= \lim_{y\to\infty}\left(\frac{y}{y-1}\right)^{y}.##
    Now let ##w = y-1##. Clearly ##y\to\infty## is equivalent to ##w\to\infty##. So
    ##\lim_{w\to\infty}\left(\frac{w+1}{w}\right)^{w+1}= \lim_{w\to\infty}\left[ \left(1+\frac{1}{w}\right)^w\left(1+\frac{1}{w}\right) \right] = e\times 1 = e## as required.
     
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