Results 1 to 8 of 8

Math Help - Limit problem (Need Help)

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    33

    Limit problem (Need Help)

    This is one of my exercise problem regarding the limit lesson.
    The problem: In the theory of relativity, the mass of a particle with velocity v is
    <br />
m =\frac{m_{0}}{\sqrt{1-v^2/c^2}}<br />
    where m_{0} is the mass of the particle at rest and c is the speed of light. What happen as  v \rightarrow c^- ?

    I am thinking that the problem is asking me to find the limit of the equation above as v approaches c from the left, but I cannot get rid of the radical sign. I try to rationalize the denominator, but it does not lead me to an answer which I think is correct.

    ANY HINT IS APPRECIATED.
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by Mathlv View Post
    This is one of my exercise problem regarding the limit lesson.
    The problem: In the theory of relativity, the mass of a particle with velocity v is
    <br />
m =\frac{m_{0}}{\sqrt{1-v^2/c^2}}<br />
    where m_{0} is the mass of the particle at rest and c is the speed of light. What happen as  v \rightarrow c^- ?

    I am thinking that the problem is asking me to find the limit of the equation above as v approaches c from the left, but I cannot get rid of the radical sign. I try to rationalize the denominator, but it does not lead me to an answer which I think is correct.

    ANY HINT IS APPRECIATED.
    Thanks
    As v approaches c the denominator tends to 0 which will make the LHS bigger.

    If you want to rationalise the denominator multiply by \sqrt{1-\frac{v^2}{c^2}}

    m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} \times \frac{\sqrt{1-\frac{v^2}{c^2}}}{\sqrt{1-\frac{v^2}{c^2}}}

    \frac{m_0\sqrt{1-\frac{v^2}{c^2}}}{1-\frac{v^2}{c^2}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by Mathlv View Post
    This is one of my exercise problem regarding the limit lesson.
    The problem: In the theory of relativity, the mass of a particle with velocity v is
    <br />
m =\frac{m_{0}}{\sqrt{1-v^2/c^2}}<br />
    where m_{0} is the mass of the particle at rest and c is the speed of light. What happen as  v \rightarrow c^- ?

    I am thinking that the problem is asking me to find the limit of the equation above as v approaches c from the left, but I cannot get rid of the radical sign. I try to rationalize the denominator, but it does not lead me to an answer which I think is correct.

    ANY HINT IS APPRECIATED.
    Thanks
    So we want \lim_{v\to{c^-}}m(v).

    Let's let u=\sqrt{1-\frac{v^2}{c^2}}.

    So, \lim_{v\to{c^-}}m(v)=\lim_{u\to0^+}m_0\left(\frac{1}{u}\right)=m  _0\lim_{u\to0^+}\frac{1}{u}=\infty.

    Does this help?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2010
    Posts
    33
    when you let the limit of m(0) as u approaches 0 from the right, is it to prevent the denominator = 0 so the function is defined? Thanks for the help.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by Mathlv View Post
    when you let the limit of m(0) as u approaches 0 from the right, is it to prevent the denominator = 0 so the function is defined? Thanks for the help.
    I'm not sure that I follow.

    tHE THING IS THAT i REALIZE TWO THINGS
    (1.) m_0 is constant
    and
    (2.) \lim_{v\to{c}^-}\sqrt{1-v^2/c^2}=0

    Now, from (2.) I realize that the denominator goes to zero, while from (1.) I realize that this will make the expression as a whole grow withought boud. So I substitute for the denominator u , and then study the behavior as u\to0^+
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jan 2010
    Posts
    33

    My Limit Problem (Understood)

    First of all thank you for showing me Vonnemo19 and e^(i*pi). I think I got it now. Here is how I understood it.

    We want to know what happen as v\rightarrow c^- in the

    \lim_{v\to c^-}\frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}

    1. since \lim_{v\to c^-}\sqrt{1-\frac{v^2}{c^2}}=\sqrt{1-\frac{c^2}{c^2}}=\sqrt{1-1}=0\Longrightarrow the denominator will get closer and closer to 0 but cannot be= 0 (else undefine).

    2. Let  u = \sqrt{1-\frac{v^2}{c^2}} and study the behavior as  u\rightarrow 0
    So:
    \lim_{v\to c^-}\frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}=  <br />
\lim_{u\to 0}\frac{m_{0}}{u}= \lim_{u\to 0}m_{0}\lim_{u\to 0}\frac{1}{u}<br />
    Since m_{0} is a constant \Longrightarrow m_{0}\lim_{u\to 0}\frac{1}{u} = \infty
    Follow Math Help Forum on Facebook and Google+

  7. #7
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by Mathlv View Post
    First of all thank you for showing me Vonnemo19 and e^(i*pi). I think I got it now. Here is how I understood it.

    We want to know what happen as v\rightarrow c^- in the

    \lim_{v\to c^-}\frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}

    1. since \lim_{v\to c^-}\sqrt{1-\frac{v^2}{c^2}}=\sqrt{1-\frac{c^2}{c^2}}=\sqrt{1-1}=0\Longrightarrow the denominator will get closer and closer to 0 but cannot be= 0 (else undefine).

    2. Let  u = \sqrt{1-\frac{v^2}{c^2}} and study the behavior as  u\rightarrow 0
    So:
    \lim_{v\to c^-}\frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}= <br />
\lim_{u\to 0}\frac{m_{0}}{u}= \lim_{u\to 0}m_{0}\lim_{u\to 0}\frac{1}{u}<br />
    Since m_{0} is a constant \Longrightarrow m_{0}\lim_{u\to 0}\frac{1}{u} = \infty
    exactly. of course there are other, more rigorous ways, but this'll do fine.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by Mathlv View Post
    when you let the limit of m(0) as u approaches 0 from the right, is it to prevent the denominator = 0 so the function is defined? Thanks for the help.
    I think that I understand what you are asking now. I did not make the substution so that we could make the function defined in any way, I did it so that I wouldn't have to deal with that big ugly denominator while trying to evaluate this particular limit.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: August 13th 2010, 01:03 AM
  2. Replies: 1
    Last Post: August 8th 2010, 11:29 AM
  3. Limit, Limit Superior, and Limit Inferior of a function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 3rd 2009, 05:05 PM
  4. another limit problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 31st 2009, 10:50 PM
  5. limit problem
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: September 13th 2008, 08:27 PM

Search Tags


/mathhelpforum @mathhelpforum