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Proving (1+x)^n approaches 1 + nx when x goes to zero

  1. May 13, 2012 #1
    1. The problem statement, all variables and given/known data
    I've heard this affirmation in an gravitation lecture. I thought it was interesting and decided to check it out. It turns out to be true so I decided to prove it.

    I'm not a mathematician so proving stuff is not my department. Still I try to do some proofs just for fun. I hope I don't make your eyes bleed or something.


    2. Relevant equations
    [itex]\lim_{x \rightarrow 0} (1 + x)^n[/itex]


    3. The attempt at a solution
    I'm not sure if I should try prooving by epsilon-delta or by mathematical induction (if that is even possible)
    I did try by mathematical induction but it didnt look convincing

    How would you do it?
     
  2. jcsd
  3. May 13, 2012 #2
    [itex]\lim_{x \to 0} (1 + x)^n [/itex]

    Apply binomial expansion to this equation and then the concept of limits.
     
  4. May 13, 2012 #3

    Curious3141

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    Try Binomial Theorem.

    While it is correct to say [itex]{(1+x)}^n \approx 1+nx[/itex] for sufficiently small x, the actual limit of the expression as x tends to zero is simply 1.
     
  5. May 13, 2012 #4

    micromass

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    It's impossible to prove this until you rigorized what the statement means.

    I think the best rigorous version of this statement is that y=1+nx is the tangent line of y=(1+x)n at the point x=0.

    Or, it could mean

    [tex]\lim_{x\rightarrow 0} \frac{(1+x)^n-(1+nx)}{x}=0[/tex]

    which means that the functions are equal up to first order.

    Try to prove these two statements.
     
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