April 4th, 2019, 04:03 AM  #31  
Senior Member Joined: Jun 2015 From: England Posts: 905 Thanks: 271  Quote:
One of the issues you need to consider in assigning a meaning to 0/0 is the deeply philosophical question Why is $\displaystyle \frac{{dy}}{{dx}} \ne \frac{0}{0}$ For instance consider Bolzano's example $\displaystyle {y^3} = a{x^2} + {a^3}$ $\displaystyle {\left( {y + dy} \right)^3} = a{\left( {x + dx} \right)^2} + {a^3}$ So $\displaystyle \frac{{dy}}{{dx}} = \frac{{2ax + adx}}{{3{y^2} + 3ydy + d{y^2}}}$ set dy, dx = 0 $\displaystyle \frac{{dy}}{{dx}} = \frac{{2ax}}{{3{y^2}}}$  
April 4th, 2019, 05:32 AM  #32 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,150 Thanks: 730 Math Focus: Physics, mathematical modelling, numerical and computational solutions  I think this example is a better indication about the thornery of division by zero. I'll use this from now on instead of the sinc x example. 
April 4th, 2019, 06:34 AM  #33 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,640 Thanks: 2624 Math Focus: Mainly analysis and algebra 
Surely the point here is that the OP proposed the idea that since $\displaystyle \lim_{x \to 0^+} \frac1x = +\infty$ and $\displaystyle \lim_{x \to 0^} \frac1x = \infty$, taking the "average" of these two "yields zero" for division by zero. Obviously, there are many problems with his reasoning, and so there are many ways to counter his argument, but one perfectly valid one is to demonstrate that there are other limits involving "division by zero" that do not result in the same value. The fact that you prefer a different rebuttal doesn't make everybody else wrong, and implying so simply exhibits an unfortunate arrogance on the part of the poster making that claim. There is space on this forum for many people to demonstrate their knowledge and understanding, each contributing to the knowledge and understanding of the OP. 

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