0

Consider the expression $\frac{2xh+3h^2}{h}$. When $h=0$ we cannot do this, as division. by zero is undefined. However, factoring the $h$ and cancelling gives $2x+3h$. Somehow, now we can substitute $h=0$ with no error at all. It seems like some cheap trick has occurred that has allowed me to bypass division by $0$.

Similarly, during first principles in calculus. When we use the limit as $h$ approaches $0$, we run into this division by $0$ error again. However, the terms cancel out, once again allowing us to evaluate the limit when we consider $h=0$.

My question, is why this is even allowed? I thought that division by $0$ is always not defined, so why can we just cancel some terms to avoid this, the reality was we still were dividing by $0$ originally. So what's going on?

Edit: After reading the linked answers, I still don't completely understand what is going on. Using my above example $\frac{2xh+3h^2}{h}=2x+3h$ is true for all $h$ except when $h=0$. I understand this. So for all values arbitrarily close to $h=0$ this is true. Hence the functions gets closer and closer to $2x$. However, it never actually reaches there, so what allows us to substitute $h=0$. We have basically said that this statement is true for all $h$ when not $0$, then have gone ahead and let $h=0$. This is extremely unintuitive to me.

James Chadwick
  • 743
  • This is a very good question, and it has also been asked many times before on this site. See https://math.stackexchange.com/questions/462199/why-does-factoring-eliminate-a-hole-in-the-limit, for instance. I have thus voted to close. If you are still left feeling confused by the linked answer, then I would suggest writing a new question that articulates what remains unclear to you. – Joe Jul 18 '23 at 11:43
  • "It seems like some cheap trick has occurred". If you want to call it a "cheap trick", then that's your problem. It is what it is. – Adam Rubinson Jul 18 '23 at 11:50
  • For your first paragraph, $\frac{2xh+3h^2}{h} = 2x + 3h\ $ if and only if $h\neq 0.$ I think this is the key point. – Adam Rubinson Jul 18 '23 at 11:53
  • Functions that are equal everywhere except at a single point have equal limits – JP McCarthy Jul 18 '23 at 22:23

1 Answers1

1

Yes, you're right. Division by $0$ is undefined, at least in the number system we're dealing in right now.

By factoring out $h$ and then substituting $0$, we aren't dividing by $0$ we are doing another thing that is not the same as division by $0$. We're doing $$\lim_{h\to0}\frac{2xh+3h^2}{h}$$ And it isn't same as division by zero.

MathStackexchangeIsNotSoBad
  • 4,716