Sinx/x uncertainty at x=0, and some confusions
Think about the function; $$\sin(x)\over x$$ We say frequently at $x=0$ this function is in a uncertain form $0\over 0$. But we conclude $$\lim_{x\to 0}\frac{sin(x)}{x}=1$$Then we say the function have no value at $x=0$ but only a limiting value at $x=0$. Hence we also say that function has a discontinuity at $x=0$ because the function's limit value is not equal to its function value at $x=0$ (at $x=0$ the function's value is uncertain).
Next i want to discuss this problem in physical viewpoint. We all know about the Fresnel diffraction law. This law states that: $$I(\theta)=I(0)\frac{sin(\beta)}{\beta}$$ Where $I(\theta)$ is the intensity of light at $\theta$ and $\beta=\beta(\theta)$. So my problem is if this function is not described at $\beta=0$ how we can say there is a bright valance at there? Another example is the <PERSON>'s energy formula: $$E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$$ For light we again have a similar situation: $$lim_{m\to 0,v\to c}\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$$ So, is there not a defined energy for light? Absolutely there is! The only solution to this mad question can be that when we are working in <PERSON>'s viewpoint, or in calculus world, we can't say just you can't divide to $0$ or $\frac{0}{0}$ is a undetermined form bla bla bla. Actualy $\frac{0}{0}$ is just the $\frac{dy}{dx}$, in our example $\frac{d(sinx)}{dx}$ and i can say it is equals to $cosx$ , then at x=0 it has the value 1! But that is the L'Hopital rule! Otherwise we can't say there is a velocity or acceleration because all of theme are actually $\frac{0}{0}$. 474
I need help.
For clarity: Think about the function $\frac{x^2-1}{x-1}$ at $x=1$ we have $\frac{0}{0}$ uncertainty again. But $\frac{x^2-1}{x-1}$=$\frac{(x-1)(x+1)}{x-1}$=$x+1$ and its value at$x=1$ is $2$. That two function is equal but one of them is uncertain at $x=1$ while second is not. And the $lim_{x\to 1}\frac{x^2-1}{x-1}=2$ too. So first of the functions has a value at $x=1$ but we just don't know what is it before limiting or manipulations. Hence there was no uncertainity at $x=1$ and the function was continuously from the beginning.
Edit: There is an assumption that the obtained function is continuous, so the limit at that point must be the same with the value at that point at physics viewpoint. Problem has been solved. 563