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Need help understanding how these limits were evaluated

  1. Jan 18, 2016 #1
    1. The problem statement, all variables and given/known data

    Hi, the problem is imply to show the following

    [tex]\lim_{n\rightarrow \infty} 10^n e^{-t} \sinh{10^{-n}t} = \lim_{n\rightarrow \infty} 10^n e^{-t} \sin{10^{-n}t} = te^{-t}[/tex]

    How can I do this? Just a hint or a first step would be great, thanks :)

    2. Relevant equations


    3. The attempt at a solution

    I've tried to make the substitution y = ln[f(x)], but I didn't get far with that.
     
  2. jcsd
  3. Jan 18, 2016 #2

    Mark44

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    I'm pretty sure they used L'Hopital's Rule after bringing out ##e^{-t}##.
     
  4. Jan 18, 2016 #3

    BvU

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    No typos ? Then you can just pull out the ##\ \ t\; e^{-t} ## !
    Do you know the $$\lim_{ y\downarrow 0}\ {\sin y\over y }\quad {\rm ?} $$
     
  5. Jan 19, 2016 #4
    That's what I ended up using :) But I'm not sure if it's ok to make the substitution n = 1/a whereby as n -> inf, then a -> 0...
     
  6. Jan 19, 2016 #5
    I also don't understand why it is works for the hyperbolic sine, but I'm happy to just accept that it does, and maybe explore the proof that it does separately; this problem isn't part of a precalculus or limits specific study effort, it appeared as part of a challenge problem in a chapter of a differential equations text I'm working on.
     
  7. Jan 19, 2016 #6

    BvU

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    For ##\sinh x \equiv {e^x-e^{-x}\over 2} \ \Rightarrow \ \sin\; i{\bf z} = i \sinh {\bf z}##
    or you can use the Taylor series of ##e^x = 1 + x + {1\over 2}x^2 + ... ## to show that
    $$\lim_{ y\downarrow 0}\; {\sinh y\over y }\quad = 1 $$
     
  8. Jan 19, 2016 #7

    Ray Vickson

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    It is much simpler than the case of ##\sin(x)/x##. Using the definition of ##\sinh(x)## we have
    [tex]\begin{array}{rcl}\sinh(x) &=&\frac{1}{2}(e^x - e^{-x}) \\
    &=& \frac{1}{2} [1 + x + x^2/2 + x^3/6 \cdots - (1 - x + x^2/2 - x^3/6 + \cdots)]\\
    &=& x + x^3/6 + \cdots
    \end{array}
    [/tex]
    so ##\sinh(x)/x = 1 + x^2/6 + \cdots \to 1## as ##x \to 0##. Alternately, just use l'Hospital's rule on ##f(x) = \sinh(x)/x##.
     
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