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.