[Archive c 17.03.2008] Humour [Archive to 28.04.2012] - page 22
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The limit exists, mathcad is right)
The root of the product is written as the product of the roots. Then one of the roots turns to zero - the whole product turns to zero.
p.s. imho, of course)
Indeed, if x tends to infinity, then by the root we have sin(0).
The limit exists, mathcad is right)
The root of the product is written as the product of the roots. Then one of the roots turns to zero - the whole product turns to zero.
p.s. imho, of course)
A little more details.
// maple 10
The trick is that the arctangent at zero is zero, and the sine is bounded. Hence the product under the root is zero.
Indeed, if x tends to infinity, then the root is sin(0).
What the hell is infinity... Time for bed.
A little more detail.
// maple 10
The trick is that the arctangent at zero is zero and the sine is bounded. Hence the product under the root is zero.
Are you saying that cos(0) / sin(0) = 0 ?
Чуть подробнее.
// maple 10
Прикол в том, что арктангенс в нуле равен нулю, а синус ограничен. Следовательно, произведение под корнем равно нулю.
Matcad also assures that there is a limit on everything, left and right. ... and in the centre... kind of a joke
PS: everything takes practice
Are you saying that cos(0) / sin(0) = 0 ?
There's not even close to that.
It's not even close.
Right, time for bed!
Gentlemen, excuse me, how do you get to to the library to the humour branch? :)
Mm-hmm...
I realize now even more how stupid I am!
and it's not humor...
lea, and Maple 10 made a mistake when going from line 4 to line 5: the limit symbol from the root can only be moved under the root if we have at least one neighborhood of the limit point (x=0), where the sub-rooted expression is positive.
For fuck's sake, people, this is the first semester of a normal technical college. Have we already forgotten that a necessary condition for the existence of the limit of a function is its definiteness in a certain neighborhood of the limit point (incidentally, it does not necessarily have to be defined at the limit point itself)?
There are heaps of theorems in all of analysis that specifically justify the correctness of limit permutations and establish their legitimacy.
The numerical example I gave Sergei is the most straightforward way to disprove the existence of a limit: I gave a sequence of values of the argument at which the subconjugate function never keeps its sign. This is enough to say that the limit does not exist.
P.S. lea, you did roughly the same thing as below:
lim( Sqrt(5-x); x->7) = Sqrt(lim(5-x); x->7) = Sqrt(-2)