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Math Help - Help with tricky limit question

  1. #1
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    Help with tricky limit question

    Here it is:
    in the theory of relativity, the mass, m of a particle with velocity v is given by:

    m = (mo) / sqrt(1 - (v^2/c^2))

    where c is the speed of light, and mo is the mass of the particle at rest. What happens to the mass of the particle as its velocity approaches the speed of light?

    we've been doing limit laws recently, so I assume im ment to use those, but this really is nothing like anything I'v seen before and im completely stuck.
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  2. #2
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    Quote Originally Posted by mrtwigx View Post
    Here it is:
    in the theory of relativity, the mass, m of a particle with velocity v is given by:

    m = (mo) / sqrt(1 - (v^2/c^2))

    where c is the speed of light, and mo is the mass of the particle at rest. What happens to the mass of the particle as its velocity approaches the speed of light?

    we've been doing limit laws recently, so I assume im ment to use those, but this really is nothing like anything I'v seen before and im completely stuck.
    You have a fraction in which the numerator is a constant and the denominator is going to 0. What happens to 1/x as x goes to 0?
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  3. #3
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    Im afraid im not quiet sure about what either of you are saying.

    Quote Originally Posted by HallsofIvy View Post
    You have a fraction in which the numerator is a constant and the denominator is going to 0. What happens to 1/x as x goes to 0?
    Well if a fraction has zero in its denominator it is usually undefined, but from my experience it will generally still have a limit, so im not quiet sure...

    Thanks for your help so far and im sory I dont inderstand...
    Last edited by mr fantastic; April 10th 2009 at 05:20 PM. Reason: Deleted reference to the deleted posts (otherwise the thread is too confusing)
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    Note that lim_{v\rightarrow c}\sqrt{1-\frac{v^2}{c^2}}=0^+ now noting this then your limit problem lim_{v\rightarrow c}\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} can be reduced to lim_{x\rightarrow0^+}\frac{m_0}{x}=m_0 lim_{x\rightarrow 0^+}\frac{1}{x} which is an easy limit.
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  5. #5
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    Quote Originally Posted by mrtwigx View Post
    Here it is:
    in the theory of relativity, the mass, m of a particle with velocity v is given by:

    m = (mo) / sqrt(1 - (v^2/c^2))

    where c is the speed of light, and mo is the mass of the particle at rest. What happens to the mass of the particle as its velocity approaches the speed of light?

    we've been doing limit laws recently, so I assume im ment to use those, but this really is nothing like anything I'v seen before and im completely stuck.
    Refer to previous posts for the mathematical reason:

    The mass of the object approaches infinity.
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