Hi!
Set P=x_{1}x_{2}\ldots x_{n} then
g\equiv \sqrt[n]{x_{1}x_{2}\ldots x_{n}} =\sqrt[n]{P}
For the real numbers w_{i}=x_{i}/g we have
w_{1}w_{2}\ldots w_{n}=(x_{1}/g)(x_{2}/g)\ldots (x_{n}/g)=P/g^n=1
So, (1) holds for wi. Then
w_{1}+w_{2}+\ldots+w_{n}\geq n (this is an...
Well, I think that the formula a^0=1 appears when you try to divide a^m by itself:
(a^m)/(a^m) = a^(m-m) = a^0
Since the first part of this equation equals 1, we have a^0=1
But if a=0 we can't do the division (0^m)/(0^m)
I'll recall another section
what is a power of a^0 ?
By definition, a^n=a*a*...a (n factors)
we can't find out what a^0 means
but a^0=1 works (if, for example, we think of (a^7)/(a^7)=1)
So we DEFINE a^0=1
The same as 0! (factorial) - we define it although there is not a...
the limit of f(x) is 1 when (x->0+) or (x->0-)
By the way f(x) can't have a value when x=0
But if we define - as you said - that f(0)=1,
then we made f(x) continuous
(left limit = right limit = f(0))
Shouldn't -ax be an integer anyway?
Ok, the limit of (1+1/n)^n is e when n ->oo and n is an integer
but what happens when we get the value
(1+1/x)^x
where x is a very large real but not an integer ??
Shouln't we prove this cases ?
I'm not quite sure about it, but I think that you should now solve these systems
x == 2 (mod5)
x == 3 (mod7)
which gives x == 17 (mod35)
and
y == 1 (mod5)
y == 4 (mod7)
which gives y == 11 (mod35)