Question

Are
f(x)=x tan[sup]-1[/sup]x& & & & & & & & (- infinity,infinty);
f(x)=x^2 tan[sup]-1[/sup]x & & & [0, infinity) ;
f(x)= cos(lnx)& & & & & & & (0,1)
couniformly continuous on the given intervals ?

Answer 1

f(x)=x tan^-1x (- infinity,infinty); as tan^-1(x) = cotan(x), it's not continuous and becomes infinitely large at the points pik, k = 0,1,2,3, x/tan(x) and not continuous at given interval.
f(x)=x² tan^-1x [0, infinity) ; as tan^-1(x) = cotan(x), it's not continuous and becomes infinitely large at the points pik, k = 0,1,2,3, x/tan(x) and not continuous at given interval.

f(x)= cos(lnx) (0,1)

f(x) is continuous on the interval (0,1)
for any two points from the interval (0,1) we can write the following expression:
|f(x1) - f(x2)| = |cos(ln(x1) - cos(ln(x2)| = |- 1/2 sin (lnx1 + lnx2)/2 sin (lnx1 - lnx2)/2 | =
= 1/2 |sin(ln(x1x2)/2) sin (ln(x1/x2)/2)| <= 1/2, so the assumption of uniformly continuity is performed at this interval.