Calculate the limit: lim(((arccos x)^(n+1))/((arccos x)^n) with x tends to 1-, for every n that belongs to N, and then prove that lim(((arccos x)^n)/(arccos x))=sprt(n).

To calculate the limit:

lim(((arccos x)^(n+1))/((arccos x)^n) with x tends to 1-, for every n that belongs to N,

We can simplify the expression by dividing both the numerator and denominator by ((arccos x)^n):

lim(((arccos x)^(n+1))/((arccos x)^n)) = lim(((arccos x)^n * arccos x)/((arccos x)^n))

Within the limit, ((arccos x)^n) in the numerator and denominator cancels out, resulting in:

lim(arccos x) as x tends to 1-

Now, to evaluate this limit, we can utilize the fact that the arccosine function is continuous. Therefore, we can replace x with the limit value, which is 1-, resulting in:

lim(arccos x) = arccos(1-)

The value of arccos(1-) is π, since arccosine is the inverse function of cosine, and at 1- (just slightly less than 1), the angle is π radians.

Now, let's move on to proving the given statement:
lim(((arccos x)^n)/(arccos x)) = sqrt(n)

We'll again simplify the expression by dividing both the numerator and the denominator by (arccos x):

lim(((arccos x)^n)/(arccos x)) = lim((arccos x)^(n-1))

Since we are taking the limit as x tends to 1-, we can replace x with the limit value:

lim((arccos x)^(n-1)) = lim((arccos 1-)^(n-1))

As mentioned earlier, arccos(1-) is equal to π. Therefore, we substitute this value into the limit:

lim((arccos 1-)^(n-1)) = lim(π^(n-1))

This is a simple limit, and we consider the exponent (n-1). As n tends to infinity, (n-1) also tends to infinity. Therefore, we have:

lim(π^(n-1)) = ∞

Hence, the given statement is incorrect. The correct limit is not equal to sqrt(n), but rather tends to infinity as n approaches infinity.