Calculus allows dividing by zero?
Wherever did you get the coefficient for i? I thought log(-10) = log(10) + pi * i = 1 + pi * i, and pi != 1,36.
ln( -10 ) = ln( 10 ) + pi * i
log( -10 ) = ln( -10 ) / ln( 10 ) = 1 + i * pi / ln( 10 ) = 1 + 1,3643763538418413474857836254314 * i
log( -10 ) = ln( -10 ) / ln( 10 ) = 1 + i * pi / ln( 10 ) = 1 + 1,3643763538418413474857836254314 * i
I get the impression that many people don''t understand them very well.
There''s nothing wrong with not understanding in my opinion. The problem comes when people think they understand.
There''s nothing wrong with not understanding in my opinion. The problem comes when people think they understand.
The most dangerous kind of fool is the one who doesn''t know he is a fool.
I think that (or something like it) is a quote from somewhere, though I have no idea where.
I think that (or something like it) is a quote from somewhere, though I have no idea where.
if there''s still some curiosity I''ll just throw s.th in:
5. (the limit as x approaches 0) x * a = x * b
//law of transitive (3,4)
6. (the limit as x approaches 0) (x * a) / x = (x * b) / x
//divide both sides of equality by the same number
The lines are NOT equivalent, because the rules applying to Limits are NOT the rules applying to normal calculus, thus:
Let lim(x->0) a(x) = a, then lim(x->0) a(x)/x has to be evaluated explicitly.
If lim(x->0) a(x) = a and lim(x->0) b(x) = b, you can calculate
lim(x->0) a(x) + b(x) = a+b and lim(x->0) a(x) b(x) = ab.
But that''s it. There are some more rules you can apply like L''Hospital, that would lead to lim x->inf x/x = 1/1 = 1, but as soon as the x changes in the equation it could be something else.
Simple Example is: lim (n->infinity) (1+1/n)^n = e
If you would do: lim (n->inf) (1+1/n) = 1, and then lim(n->inf) 1^n = 1, you''d get something totally else.
5. (the limit as x approaches 0) x * a = x * b
//law of transitive (3,4)
6. (the limit as x approaches 0) (x * a) / x = (x * b) / x
//divide both sides of equality by the same number
The lines are NOT equivalent, because the rules applying to Limits are NOT the rules applying to normal calculus, thus:
Let lim(x->0) a(x) = a, then lim(x->0) a(x)/x has to be evaluated explicitly.
If lim(x->0) a(x) = a and lim(x->0) b(x) = b, you can calculate
lim(x->0) a(x) + b(x) = a+b and lim(x->0) a(x) b(x) = ab.
But that''s it. There are some more rules you can apply like L''Hospital, that would lead to lim x->inf x/x = 1/1 = 1, but as soon as the x changes in the equation it could be something else.
Simple Example is: lim (n->infinity) (1+1/n)^n = e
If you would do: lim (n->inf) (1+1/n) = 1, and then lim(n->inf) 1^n = 1, you''d get something totally else.
quote:Original post by thedustbustr
I realized that by using limits you can divide by zero.
You can never divide anything by zero. You are basically stating that you can divide something with nothing and get eternity. This is not possible by any means. Nor can you divide nothing with something and get something.
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