Question Statement

According to Einstein's theory of relativity, the mass $m$ of a body moving with velocity $v$ is given by:

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

where $m_0$ is the initial mass (rest mass) and $c$ is the speed of light.

What happens to $m$ as $v\to c^{-}$ (as $v$ approaches $c$ from below)?

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Background and Explanation

This problem explores the concept of limits at infinity, specifically what happens when a denominator approaches zero while the numerator remains constant. Understanding one-sided limits is crucial here since the velocity $v$ cannot exceed $c$, so we only consider the approach from below ($v<c$).

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Solution

Method 1: Conceptual Analysis

We analyze the behavior of the mass formula as velocity increases toward the speed of light:

As $v$ increases and gets closer to $c$:

• The ratio $\frac{v^2}{c^2}$ approaches $1$: $\frac{v^2}{c^2}\to 1$
• Therefore, the expression inside the square root approaches zero: $1-\frac{v^2}{c^2}\to 0$
• Taking the square root, the denominator approaches zero: $\sqrt{1-\frac{v^2}{c^2}}\to 0$
• Since the numerator $m_0$ remains constant, we have a constant divided by an increasingly small positive number: $m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\to \infty$

Hence, when $v\to c^{-}$, then $m\to \infty$.

Method 2: Formal Limit Evaluation

Given the mass formula: $m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$

Taking the limit as $v\to c$:

$\begin{array}{rl}\underset{v\to c}{\lim }m& =\underset{v\to c}{\lim }\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\\ & =\frac{m_0}{\sqrt{1-\frac{c^2}{c^2}}}\text{(substituting }v=c\text{)}\\ & =\frac{m_0}{\sqrt{1-1}}\\ & =\frac{m_0}{\sqrt{0}}\\ & =\frac{m_0}{0}\\ & =\infty \end{array}$

This shows that when $v\to c$, then $m\to \infty$.

Additional Limit Calculations (Separate Problem)

The following content appears to be from a different problem involving piecewise function continuity at $x=3$:

Left Hand Limit (L.H.L):

$\begin{array}{rl}\underset{x\to 3^{-}}{\lim }f(x)& =\underset{x\to 3^{-}}{\lim }(kx+1)\\ & =k(3)+1\\ \underset{x\to 3^{-}}{\lim }f(x)& =3k+1\end{array}$

Right Hand Limit (R.H.L):

$\begin{array}{rl}\underset{x\to 3^{+}}{\lim }f(x)& =\underset{x\to 3^{+}}{\lim }(2-kx)\\ & =2-k(3)\\ \underset{x\to 3^{+}}{\lim }f(x)& =2-3k\end{array}$

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Key Formulas or Methods Used

• Relativistic mass formula: $m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$
• Limit of a quotient: When ${\lim }_{x\to a}f(x)=L\ne 0$ and ${\lim }_{x\to a}g(x)=0$, then ${\lim }_{x\to a}\frac{f(x)}{g(x)}=\infty$ (sign depends on direction of approach)
• Direct substitution property: For continuous functions, ${\lim }_{x\to a}f(x)=f(a)$
• One-sided limits: Analysis of $v\to c^{-}$ ensures we approach from the physical domain where $v<c$

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Summary of Steps

1. Identify the relativistic mass formula $m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$
2. Analyze the denominator's behavior as $v\to c^{-}$: observe that $\frac{v^2}{c^2}\to 1$
3. Determine that $1-\frac{v^2}{c^2}\to 0^{+}$ (approaches zero through positive values)
4. Recognize that the square root $\sqrt{1-\frac{v^2}{c^2}}\to 0$
5. Conclude that dividing constant $m_0$ by an infinitesimally small positive number yields $m\to \infty$
6. (Alternative) Apply formal limit notation: substitute $v=c$ to obtain $\frac{m_0}{0}=\infty$