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Help w/ limit

  1. Dec 9, 2004 #1
    [tex]\lim_{x\rightarrow\zero} \frac{\sin(x)}{\exp(x)}[/tex]
     
    Last edited: Dec 9, 2004
  2. jcsd
  3. Dec 9, 2004 #2

    quasar987

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    Why wouldn't that be simply 0 ?
     
  4. Dec 9, 2004 #3
    That's what i'm wondering. I have been trying to solve this limit for about 20 minutes now. BTW, that is the limit as x goes to 0... I don't know how to get that into the latext graphic :/
     
  5. Dec 9, 2004 #4

    quasar987

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    As you know, in Latex, the \ "operator" is only used to input a special comand.. so you don't need one before writing 0 because zero is just a number. So it's simply

    lim_{x \rightarrow 0} :smile:

    As for the justification of the limit, we have that the limit of a quotient is the quotient of the limit if the limit exists and if the function at the denominator is never zero at least past a certain aribitrarily large x.

    the limit of sinx is sin0 = 0 since sin x is continuous over all real.

    the limit of exp(x) is e^0 =1 also because e^x is continuous on all real.

    Therefor the limit of the quotient is 0/1 = 0.
     
    Last edited: Dec 9, 2004
  6. Dec 9, 2004 #5

    quasar987

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    A cool proof for the limit of sinx makes use of the identity [itex]0 \leq |sinx| \leq |x|[/itex] [itex]\forall x \in \mathbb{R}[/itex] because you then have that

    [tex]\lim_{x \rightarrow 0} 0 = 0[/tex]

    and

    [tex]\lim_{x \rightarrow 0} |x| = 0[/tex].

    such that, by the sandwich theorem,

    [tex]\lim_{x \rightarrow 0} |sinx| = 0[/tex]

    And also using the "theorem" according to which

    [tex]\lim_{x \rightarrow x_0} f(x) = 0 \Leftrightarrow \lim_{x \rightarrow x_0} |f(x)| = 0[/tex]

    (as you can very easily prove using the epsilon-delta definition of limit), we find the answer:

    [tex]|sinx| \rightarrow 0 \Rightarrow sinx \rightarrow 0 [/tex]
     
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