# Thread: Limit problem (Need Help)

1. ## 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
$\displaystyle m =\frac{m_{0}}{\sqrt{1-v^2/c^2}}$
where $\displaystyle m_{0}$ is the mass of the particle at rest and c is the speed of light. What happen as $\displaystyle v \rightarrow c^-$ ?

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

2. Originally Posted by Mathlv
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
$\displaystyle m =\frac{m_{0}}{\sqrt{1-v^2/c^2}}$
where $\displaystyle m_{0}$ is the mass of the particle at rest and c is the speed of light. What happen as $\displaystyle v \rightarrow c^-$ ?

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 $\displaystyle \sqrt{1-\frac{v^2}{c^2}}$

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

$\displaystyle \frac{m_0\sqrt{1-\frac{v^2}{c^2}}}{1-\frac{v^2}{c^2}}$

3. Originally Posted by Mathlv
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
$\displaystyle m =\frac{m_{0}}{\sqrt{1-v^2/c^2}}$
where $\displaystyle m_{0}$ is the mass of the particle at rest and c is the speed of light. What happen as $\displaystyle v \rightarrow c^-$ ?

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 $\displaystyle \lim_{v\to{c^-}}m(v)$.

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

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

Does this help?

4. 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.

5. Originally Posted by Mathlv
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.) $\displaystyle m_0$ is constant
and
(2.) $\displaystyle \lim_{v\to{c}^-}\sqrt{1-v^2/c^2}=0$

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 $\displaystyle u$ , and then study the behavior as $\displaystyle u\to0^+$

6. ## 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 $\displaystyle v\rightarrow c^-$ in the

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

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

2. Let $\displaystyle u = \sqrt{1-\frac{v^2}{c^2}}$ and study the behavior as $\displaystyle u\rightarrow 0$
So:
$\displaystyle \lim_{v\to c^-}\frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}= \lim_{u\to 0}\frac{m_{0}}{u}= \lim_{u\to 0}m_{0}\lim_{u\to 0}\frac{1}{u}$
Since $\displaystyle m_{0}$ is a constant $\displaystyle \Longrightarrow m_{0}\lim_{u\to 0}\frac{1}{u} = \infty$

7. Originally Posted by Mathlv
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 $\displaystyle v\rightarrow c^-$ in the

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

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

2. Let $\displaystyle u = \sqrt{1-\frac{v^2}{c^2}}$ and study the behavior as $\displaystyle u\rightarrow 0$
So:
$\displaystyle \lim_{v\to c^-}\frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}= \lim_{u\to 0}\frac{m_{0}}{u}= \lim_{u\to 0}m_{0}\lim_{u\to 0}\frac{1}{u}$
Since $\displaystyle m_{0}$ is a constant $\displaystyle \Longrightarrow m_{0}\lim_{u\to 0}\frac{1}{u} = \infty$
exactly. of course there are other, more rigorous ways, but this'll do fine.

8. Originally Posted by Mathlv
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.