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Some special consequences from the limits of e: need a little help

  1. Jan 21, 2005 #1
    Well, hello!
    I have difficuties solving a problem.
    Well, the first 3 ones are done successfully:
    a. Calculate out the lim (x -> inf) of (1+2/x)^x
    But assuming that 2/x = 2/t, I got the result to be e^2
    b. Do the same thing with the lim (x ->0) of (1+3x)^(1/x)
    Well, yeah "do the same thing", I assumed that t = 3x; I got the result of e^3
    c.lim (x->0) of [(e^3x)-1]/x . Well, t = 3x this time. I got it as 3
    But the third: d
    lim (x->0) of { ln(1+sinx)/x }
    Well, I tried assuming that t = sinx , but I got stuck in solving this, so far. Please give me a suggestion! Thanks!
     
  2. jcsd
  3. Jan 21, 2005 #2

    dextercioby

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    Did u study derivatives and L'Hôspital's law??

    Daniel.

    P.S.The first 3 are perfect.
     
  4. Jan 21, 2005 #3

    dextercioby

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    There are 2 approximations which u can use:

    For "x" very,very small
    [tex] \sin x\sim x [/tex]
    [tex] e^{x}\sim 1+x [/tex]

    Use these 2 approximations for good purpose...

    Daniel.
     
  5. Jan 21, 2005 #4
    No I haven't learnt about it yet. Any other way?
     
  6. Jan 21, 2005 #5

    dextercioby

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    Use the approximations.The first one can be proven via intuitive geometrical resoning and the second,well,it's a bit difficult...

    Daniel.

    P.S.The final result is "+1".
     
  7. Jan 21, 2005 #6

    HallsofIvy

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    Changing x to t (as in 2/x= 2/t) doesn't do anything! If you have learned that lim(t->infinity) (1+ 1/t)t= 2 then what you can do is let t= x/2. I'll bet that's what you actually did. Of course, then x= 2t and, clearly, as x-> inf, t-> inf. The limit becomes
    lim(t->inf) (1+1/t)2t= lim(t->inf) ((1+ 1/t)[sup t[/sup])2 which, since x2 is continuous, (is lim(t->inf) (1+1/t)t)2= e2.
    I assume that "(1+3x)" was really (1+ 3/x). If so, yes, that's correct.
    Is that e2x or e3x? I assume the latter. Taking t= 3x, x= t/3 so this become lim(t->0)3(et-1)/t. (et-1)/t goes to 1 because it is the derivative of ex at x= 0?
    How about using the fact that ln(1+ sin x)/x= ln{(1+ sin x)1/x}. Does that help? If we let y= ln{(1+sinx)1/x}, then
    ey= (1+ sin x)1/x. Can you find the limit of that?
     
  8. Jan 21, 2005 #7
    Ok, I got it! I was so stupid!
    Thanks, Mr. Hallsofivy, thanks Mr. Dextercoby
     
    Last edited: Jan 21, 2005
  9. Jan 21, 2005 #8
    Sorry, misters, but I got stuck while finding the lim of e^y.
     
  10. Jan 21, 2005 #9

    dextercioby

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    Halls said that
    [tex] \lim_{x\rigtarrow 0} \frac{1}{x}\ln(1+\sin x)=\lim_{x\rightarrow 0} [\ln(1+\sin x)]^{\frac{1}{x}} [/tex]

    Denote the limit by "L" and exponentiate the previous relation.That's what Halls said.
    [tex] e^{L}=\lim_{x\rightarrow 0}(1+\sin x)^{\frac{1}{x}} [/tex]

    And now use the trick:
    [tex] \lim_{x\rightarrow 0} (1+x)^{\frac{1}{x}}=\lim_{\frac{1}{x}\rightarrow +\infty}(1+\frac{1}{\frac{1}{x}})^{\frac{1}{x}}=e [/tex]

    and the fact that
    [tex] \lim_{x\rightarrow 0}\frac{\sin x}{x}=1 [/tex]

    to find your answer.

    Daniel.
     
  11. Jan 21, 2005 #10
    Why [tex] e^{L}=\lim_{x\rightarrow 0}(1+\sin x)^{\frac{1}{x}} [/tex] ?
     
  12. Jan 21, 2005 #11

    dextercioby

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    Well,since:
    [tex] L=\lim_{x\rightarrow 0} \ln(1+\sin x)^{\frac{1}{x}} [/tex]

    ,what do you get by exponentiating the equation?

    Daniel.
     
  13. Jan 21, 2005 #12
    Well, I got it as lim (x->0) of [(1+sinx)^(1/x)]
     
  14. Jan 21, 2005 #13

    dextercioby

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    Okay now use the trick with the reversing fractions and the of sine ratio.

    Daniel.
     
  15. Jan 21, 2005 #14
    Okay, thanks!
    PS: Thanks for telling me about LaTex! It's cool! I'm learning it!
     
  16. Jan 21, 2005 #15
    Hmmm... So far, I've done it. But I think tis way is better:

    [tex]\lim_{x\rightarrow0} \frac{ln(1+sinx)}{x}[/tex]

    =[tex]\lim_{x\rightarrow0} \frac{ln(1+sinx)}{sinx} . \frac{sinx}{x}[/tex]

    =[tex]1.1 = 1[/tex]

    Is it right?
     
    Last edited: Jan 21, 2005
  17. Jan 21, 2005 #16

    dextercioby

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    Nope.Can u prove that the first limit is "+1"??I frankly doubt it.

    BTW,[tex] \sin x [/tex]
    The functions should not be italic...


    Daniel.
     
  18. Jan 22, 2005 #17
    If you have two functions which are both tending to zero as x tends to zero then they can be both replaced by equivalent infinitesimals. The approximation, [itex]\sin x = x[/itex] near zero, would be useful here (though not without justification) if you do not know that [itex]\frac{ln(1+x)}{x}[/itex] tends to 1 as x tends to zero or you are precluded from using the first form L'Hopital's Theorem.
     
  19. Jan 22, 2005 #18
    Uhmm... alright!
    Just imagine that [tex]sinx = \alpha[/tex]
    Then the thing above is equal to:

    [tex]\lim_{x\rightarrow0}\frac{ln(1+\alpha)}{\alpha}[/tex] (*)

    Then, this limit is a consequence limit of e:

    [tex]\lim_{x\rightarrow\infty}(1+\frac{1}{x})^x = e[/tex]

    So. (*) =1
     
  20. Jan 22, 2005 #19
    Hmm... yeah, the basic consequence
    [tex]\lim_{x\rightarrow0}\frac{ln(1+u)}{u}=1[/tex]
     
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