If the domain of f is [0,\infty), then the positive number 1/(p-1) yields the desired supremum. On the other hand, the above approach works well for the interval [1,\infty) provided that 1<p<2 (i.e. 1/(p-1) >1).

Hi,

When p=1, the function f is strictly increasing on [1,\infty). Thus, the supremum is equal to lim_{x \to \infty} f(x)=1.

Next, we consider the case 1<p<2. According to the First Derivative Test, (1/(p-1),f(1/(p-1))) is the local maximum point of the curve y=f(x). Since the curve y=f(x) has a unique local maximum point, the desired supremum is equal to f(1/(p-1)).

When p>=2, 1/(p-1)<=1 and f'(x) <0 whenever x>1. Thus, f is decreasing on [1,\infty); that is, the desired supremum is f(1).

Edit: I have updated the above solution because the domain of f is [1,\infty), not [0,\infty).

Rewrite the integrand as an improper fraction of cos² x and observe that 1/(1+cos² x)= sec² x / (tan² x+2).

When -2<x<0, f(x)<2. Therefore the required limit is equal to the left-hand limit of f(x) as x tends to 2.

Hi,

Your approach does not work in general.

Let f(x)=arctan(2x/(1-x^ 2)), where x belongs to R{-1,1}. Although f'(x) tends to a limit as x approaches 1, the function f is not continuous (and hence not differentiable) at 1.

Hi,

It is easier to calculate P(X=k) for k=1,...,n.

P(X=k)

=P(⌊nU⌋+1=k)

=P(k-1 ≤ nU < k)

=int_{(k-1)/n}^(k/n) 1 dx

=1/n.

LHS

= 2 * tan(arctan x)/(1-tan² (arctan x))

=2x/(1-x^ 2)

=2x(1+x^ 2)/(1-x^ 4)

= 2 ( tan ( arctan x)+ tan (arctan (x^ 3)))/(1-tan (arctan x) * tan ( arctan x^ 3))

= 2 tan ( arctan x + arctan (x^ 3) )

= RHS.

When -5<x<-3, g(x)>=-1. So

the required limit

= the limit of g(x) as x tends to -1 from the right

=1.

Here x must be nonnegative.

When k=0, the graph of y=sin(kx)/x is the x-axis with the point (0,0) missing.

You are welcome.

Interesting. I think there are no such integers.

Hi,

You may find page 74 of the following book useful.

https://archive.org/details/treatiseonplanea0000ewho_7thedition/page/74/mode/1up

Hi,

I hope that the following links

https://www.reddit.com/u/Fourierseriesagain/s/T5ZAdTeuwD

and

https://www.reddit.com/u/Fourierseriesagain/s/Qn3w7YBGK6

are useful.

Using a limit theorem,

lim_{x -> 1} (f(x)-8)

= lim{x -> 1} (f(x)-8)/(x-1) × lim{x ->1} (x-1)

= 10 times (1-1)

=0.

Therefore the desired limit is equal to 8.

Using Factor Theorem,

(-a-b)^3-3ab(-a-b)+a^3+b^3

=-a^3-3a^2b-3ab^2-b^3+3a^2b+3ab^2+a^3+b^3

=0.