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Limit proof problem as x goes to infinity

  1. Nov 11, 2013 #1
    1. The problem statement, all variables and given/known data
    prove that ##\lim_{x \to \infty} \frac{\sqrt{x+1}}{x} = 0## where ##x>0##



    2. Relevant equations

    Definition: Let ##A\subseteq\mathbb{R}## and let ##f:A\rightarrow \mathbb{R}##. Suppose that ##(a,\infty)\subseteq A## for some ##a\in\mathbb{R}##. We say that ##L\in\mathbb{R}## is the limit of ##f## as ##x\rightarrow\infty## and write ##\lim_{x \to \infty} f=L## if any given ##\epsilon>0## there exists a ##K=K(\epsilon)>a## such that for any ##x>K## then ##|f(x)-L|<\epsilon##.

    3. The attempt at a solution
    I can't seem to connect this definition to the problem because the part that is confusing is connecting ##a## to ##\epsilon##. What I have so far is Suppose that ##(1,\infty)\subseteq A## where ##a=1##. Let ##\epsilon >0## then there exists a ##K\in\mathbb{N}## such that if ##x>K## then... Let's just say im completely stuck.
     
    Last edited: Nov 11, 2013
  2. jcsd
  3. Nov 11, 2013 #2

    tiny-tim

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    hi bonfire09! :smile:
    you hae to prove that if x > a, then (√(x+1))/x < ε :wink:
     
  4. Nov 11, 2013 #3
    Yeah but what do I let ##\epsilon## equal to? Do I let it be equal to ##a##? Thats the part im having trouble with. yes if ##x>a## then I can't really show ##\frac{\sqrt{x+1}}{x}<\epsilon##.
     
    Last edited: Nov 11, 2013
  5. Nov 11, 2013 #4

    tiny-tim

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    ahh! …

    no, a is a function of ε :wink:

    (in these problems, it's usually something like 1/ε or 1/√ε or 1/ε2)
     
  6. Nov 11, 2013 #5
    Well it looks similar to the normal epsilon proofs for limits of sequences. Here is what I have for scratchwork. We want ##|f(x)-0|<\epsilon \implies |\frac{\sqrt{x+1}}{x}|=\frac{\sqrt{x+1}}{x}\leq \sqrt{x+1}\leq x+1 <\epsilon## for ##x>0##. But then i am not sure what I should set my epsilon too?

    Do I got about this way then? Let ##\epsilon>1\implies \epsilon-1>0## and let ##a=\epsilon-1## and there exists a ##k>\epsilon-1## such that if ##x>k##.... Would that be the way I should do it? THough I am not sure if that epsilon will work.
     
    Last edited: Nov 11, 2013
  7. Nov 12, 2013 #6

    tiny-tim

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    no, ε has to be very close to 0
     
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