Why Newton’s Calculus Not Rigorous?
$latex f(x ) = \frac {x(x^2+ 5)} { x}$ ...[1]
cancel x (≠0)from upper and below => $latex f(x )=x^2 +5 $
$latex \mathop {\lim }\limits_{x \to 0} f(x) =x^2 +5= L=5 $ ...[2]
In [1]: we assume x ≠ 0, so cancel upper & lower x
But In [2]: assume x=0 to get L=5
[1] (x ≠ 0) contradicts with [2] (x = 0)
This is the weakness of Newtonian Calculus, made rigorous later by Cauchy's ε-δ 'Analysis'.
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