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Limits - proving their equality

  1. Apr 21, 2007 #1
    Hello!
    All of us know, that the constant e is defined as:
    e = lim[n->oo] (1+ 1/n)^n

    I'm proving the derivative (ln(x))'=1/x and at some point I have to show that the limit:
    lim[n->0] (1 + n)^(1/n)
    is equal to the one previously mentioned, the definition of e. Probably I will have to think separately about n->+oo and n->-oo, but... There is still one, the most obvious question:
    How can I do that? What should I start with? Or maybe it is impossible to be proved algebraicaly and all I can do is just input that limit into the calculator and see the limit?

    Thank you very much for your help!
    Greetings,
    Theriel
     
  2. jcsd
  3. Apr 21, 2007 #2

    Gib Z

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    Actually you are making it much harder than it really is, thank god :)

    In the definition you posted, the limit is an n tends to infinity. The 1/n here therefore tends to 0. And the exponent, tends to infinity, obviously.

    In the Other limit, it is just showing the same thing. As n goes to zero, the term after the 1 still approaches the same thing. And the exponent also approaches the same thing.

    In general this is a neat trick, if we have a limit of n tending to infinity, replace n with 1/n, and make the n approach zero and they are equal.
     
  4. Apr 21, 2007 #3

    HallsofIvy

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    Or, if it makes more sense to you, replace n by 1/x. As n goes to 0, 1/x goes to infinity.
     
  5. Apr 21, 2007 #4
    It looks too easy as for my maths professor so I am almost sure that there was something I forgot about ;-]. Nevertheless - once again, thank you for your advice and time!
     
  6. Apr 25, 2007 #5
    Heh, yeah, I KNEW that there would be a problem... so:

    During our classes we defined "e" as:
    e = lim[n->+oo] (1+ 1/n)^n

    Now I have to prove (algebraically) that the limit for n tending to -oo (negative infinity) is also equal exactly e. And it is not enough to state simply that it works or I saw a graph etc... it must be a mathematical proof.

    I would highly appreciate your any ideas ;-].

    Thank you for help!
     
  7. Apr 25, 2007 #6

    Dick

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    Take the log of the expression and prove the limit is 1 using l'Hopital.
     
  8. Apr 25, 2007 #7
    I am not very familiar with l'Hopital theorem, so I am basing on wiki and planetmath (hopefully properly understood)

    Here, according to your hint, I must prove:
    lim[n->-oo] ln((1+ 1/n)^n) = 1

    so I have (n->-oo):
    ln ( lim(n+1)/lim(n) ) * lim(n) = 1

    here, using the l'Hopital theorem (limit of f/g = limit of f'/g'):

    ln (1) * lim (n) = 1
    0 * (-infinity) = 1

    OK, it does not look easier ;-]. Zero times infinity is undefined... Was that meant to be done this way or you thought about some other approach?
     
  9. Apr 25, 2007 #8

    Dick

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    You can only use l'Hopital for 0/0 and infinity/infinity type cases. Instead of n*ln(1+1/n) - write it as ln(1+1/n)/(1/n). Now its 0/0. Now use l'Hopital.
     
  10. Apr 25, 2007 #9
    So we have a problem #-/ When we have:
    ln(1+1/n)/(1/n)

    we have to differentiate ln(1+1/n). So we have to use the formula ln(x)' = 1/x. And this formula is to be proved ;-D.

    Just to point it out -> my MAIN task was to prove (e^x)' = e^x. I did it using ln(x)'. Then I had to prove ln(x)' (because during our classes we proved it using e^x). I did it, however presenting e^x in different form (see previous posts). Now, I am just to prove it for -oo...

    That is why I cannot differentiate that... unless I made some logical mistake...
     
  11. Apr 25, 2007 #10

    Dick

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    You don't need the limit n->-infinity to prove ln'(x)=1/x. You said you'd already proved that. So I was hoping you'd be ok with just using the derivative formula now.
     
  12. Apr 25, 2007 #11
    No no no.... maybe I would post you my whole problem, it will be easier to find some solution? By the way, thank you for your help and time, sincerely....

    1.I proved e^x using ln(x)', chain rule, (x)'
    2.I had to prove ln(x)', the final step:

    1/x * ln (lim [n->0] (1+n) ^ (1/n) ) ---by def. of e --- 1/x * ln(e) = 1/x
    hence, ln(x)' = 1/x

    To make the proof ln(x) complete I was to prove that lim [n->0] (1+n) ^ (1/n) is equal to e.

    During our classes we defined "e" as:
    e = lim[n->+oo] (1+ 1/n)^n

    Changing the first formula into:
    e = lim[n->oo] (1+ 1/n)^n
    wasn't a big problem. However, there was one difference between my definition and the one we learned about. So I am to prove that the limit holds for n->-oo ;-].

    I hope I explained my problem clearly... I am proving a few things in a chain and that is why I cannot use any of them at any steps (hence no derivative of e^x, ln(x) and so on)
     
    Last edited: Apr 25, 2007
  13. Apr 25, 2007 #12

    Dick

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    You are welcome for the help! But I'm still not quite catching at what point you need to prove the n->-infinity part. What is there about your proof of the derivative property that makes you need both signs?
     
  14. Apr 25, 2007 #13
    My proof assumes that e=lim[n->oo] (1+ 1/n)^n (hence, for n tending to both +,- infinity) And, theoretically, I only know that e=lim[n->+oo] (1+ 1/n)^n . And that makes the problem ;-].
     
  15. Apr 25, 2007 #14

    VietDao29

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    Errr...
    So, what you'd like to ask is to use:
    [tex]\lim_{x \rightarrow \fbox{+ \infty}} \left(1 + \frac{1}{x} \right) ^ x = e[/tex]
    to prove that:
    [tex]\lim_{x \rightarrow \fbox{- \infty}} \left(1 + \frac{1}{x} \right) ^ x = e[/tex], right?

    Ok, so let t = -x, so [tex]x \rightarrow - \infty \Rightarrow t \rightarrow + \infty[/tex], the whole expression be comes:

    [tex]\lim_{x \rightarrow \fbox{- \infty}} \left(1 + \frac{1}{x} \right) ^ x = \lim_{t \rightarrow \fbox{+ \infty}} \left(1 - \frac{1}{t} \right) ^ {-t} = \lim_{t \rightarrow \fbox{+ \infty}} \frac{1}{\left(1 - \frac{1}{t} \right) ^ {t}}[/tex]

    [tex]= \lim_{t \rightarrow + \infty} \frac{\left(1 + \frac{1}{t} \right) ^ {t}}{\left(1 - \frac{1}{t} \right) ^ {t} \left(1 + \frac{1}{t} \right) ^ {t}}[/tex]

    [tex]= \lim_{t \rightarrow + \infty} \left( \frac{1 + \frac{1}{t}}{1 - \frac{1}{t}} \right) ^ t \frac{1}{\left(1 + \frac{1}{t} \right) ^ {t}}[/tex]

    [tex]= \frac{1}{e} \lim_{t \rightarrow + \infty} \left( \frac{1 - \frac{1}{t} + \frac{2}{t}}{1 - \frac{1}{t}} \right) ^ t[/tex]

    [tex]= \frac{1}{e} \lim_{t \rightarrow + \infty} \left( 1 + \frac{\frac{2}{t}}{1 - \frac{1}{t}} \right) ^ t[/tex]

    [tex]= \frac{1}{e} \lim_{t \rightarrow + \infty} \left( 1 + \frac{2}{t - 1} \right) ^ t[/tex]

    [tex]= \frac{1}{e} \lim_{t \rightarrow + \infty} \left( 1 + \frac{1}{\frac{t - 1}{2}} \right) ^ t = ...[/tex]

    You should be able to go from here, right? :)
     
    Last edited: Apr 25, 2007
  16. Apr 26, 2007 #15
    Yeah, I should be able... but ;-]
    I think I have missed something or there is something basic I cannot think about...

    In your calculations we have to prove now that the limit is equal to e^2 (to make the whole result equal to e). We may:
    - use t=-x again, however (after having done some calculations) it gives us nothing.
    -make the same denominator, which gives us:
    1/e * lim [t->oo] ((t+1)/(t-1))^t.
     
    Last edited: Apr 26, 2007
  17. Apr 26, 2007 #16

    VietDao29

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    Nope, why should you make the substitution t = -x? When [tex]t \rightarrow + \infty \Rightarrow x \rightarrow - \infty[/tex], and you cannot use the limit: [tex]\lim_{\alpha \rightarrow + \infty} \left( 1 + \frac{1}{\alpha} \right) ^ \alpha = e[/tex] to complete your problem.
    The limit is not 1. You should note that [tex]1 ^ \infty \neq 1[/tex], it's one of the Indeterminate Forms.

    Ok, big hint of the day. :)

    We have:
    [tex]\lim_{\alpha \rightarrow + \infty} \left( 1 + \frac{1}{\alpha} \right) ^ \alpha = e[/tex]
    And, we also have that:
    [tex]\lim_{x \rightarrow - \infty} \left(1 + \frac{1}{x} \right) ^ x = \lim_{t \rightarrow + \infty} \left(1 - \frac{1}{t} \right) ^ {-t} = \frac{1}{e} \lim_{t \rightarrow + \infty} \left( 1 + \frac{1}{\frac{t - 1}{2}} \right) ^ t[/tex]

    Now, if you let [tex]\alpha = \frac{t - 1}{2}[/tex], then when [tex]t \rightarrow + \infty[/tex], you also have: [tex]\alpha \rightarrow + \infty[/tex], right?

    Now, do some little manipulation, and change t to [tex]\alpha[/tex], you'll arrive at the correct result in no time. :smile:
     
  18. Apr 26, 2007 #17

    Dick

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    Try this argument on for size. According to my reference e^x is defined as lim[n->infinity](1+x/n)^n.

    So 1/e=lim[n->infinity](1-1/n)^n=lim[n->-infinity](1+1/n)^(-n)=
    1/lim[n->-infinity](1+1/n)^n.

    So 1/e is equal to 1/(your limit). So your limit=e.
     
  19. Apr 26, 2007 #18
    Yeah, I forgot about ^n in the previous approach.

    Thank you very much for your help and hint! ;-] Soo... I have:
    e * lim[A->+oo] (1 + 1/A)

    Is the limit of 1/inifinity indiscussably equal to zero or I may expect another task from my teacher to prove it ? ;-D.
     
  20. Apr 26, 2007 #19

    VietDao29

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    Err... He only knows that:
    [tex]\lim_{n \rightarrow + \infty} \left( 1 + \frac{1}{n} \right) ^ n = e[/tex], we hasn't covered what ex is... :rolleyes:
     
  21. Apr 26, 2007 #20

    VietDao29

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    Hurray, you get it correctly. The limit: [tex]\lim_{x \rightarrow \infty} \frac{1}{x} = 0[/tex] is well-know, hence, you don't have to "re-prove" it.
    Congratulations. :smile:
     
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