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Confusion about the Heaviside method

  • Thread starter ali PMPAINT
  • Start date
Problem Statement
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Relevant Equations
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So, Heavside's method confused me, I mean, but we can't divide by zero, can we?
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So, my mind sees it as cheating, and to multiply both sides by zero to cancel zeros, and for the example above(from Thomas Calulus), when x = 1, before you multiply by x-1(=0), you get: 2/0=A/0+B/-1+c/-2, and I think you get the point where I am confused, where am I misunderstanding Heavside's method?
 
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But it is not multiplied by zero. It is multiplied by the variable ##x-1##.
Either case, before or after canceling ##x-1##. The equality holds for the limit ##x=1##.

You cannot cancel zeros, but you can cancel expressions, because same expressions have the same behavior (Same value at any chosen point). Take the following:

$$\lim_{x\rightarrow 0} \frac{x}{x}=1$$

Even without cancelling the terms ##x##, the limit is ##1##. Although, approaching the function ##x## itself is zero. Cancelling expressions is not the same as cancelling numbers.

You can check the previous with a simple table, approaching zero from the left and the right of ##x##.

**Not an expert, but this is what I think.
 
You cannot cancel zeros, but you can cancel expressions, because same expressions have the same behavior (Same value at any chosen point). Take the following:

$$\lim_{x\rightarrow 0} \frac{x}{x}=1$$
Well, you're kind of right, but $$\lim_{x\rightarrow 0} \frac{x}{x}=1$$ doesn't imply x/x = 1 when x = 0, does it?
So, if you're saying that we should take the limit as x is approaching to a specific number, like 1 on the first post example, it doesn't mean that the value EQUALS the number it approaches, so, although I liked your reasoning, I'm still not convinced.
 
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\lim_{x\rightarrow 0} \frac{x}{x}=1 doesn't imply x/x = 1 when x = 0, does it?
The limit is 1 for ##x\rightarrow 0##.

Well, it is more like a removable discontinuity. Sadly, this is my best explanation, maybe others can help us understand.
 

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