290- 3 If x=0, (nx)/(1+nx) = 0 for all n, hence in the limit. If x .gt. 0, nx goes to infinity, hence the ratio goes to 1 11 If x .leq. a (and non-negative), x/(x+n) .leq. a/(a+n) [take the derivative of x/(x+n) if you want to show this]. Since a/(a+n) .lt. epsilon if n .gt. a/epsilon, convergence to 0 is uniform. x/(x+n) .lt. epsilon requires that n .gt. x/epsilon - x, which for small epsilon blows up with x, hence convergence to 0 is not uniform. 21 There exists K(epsilon/2) such that |f_n - f| .lt. epsilon/2 if n .gt. K(epsilon/2) There exists L(epsilon/2) such that |g_n - g| .lt. epsilon/2 if n .gt. L(epsilon/2) Therefore, |f_n - f + g_n - g| .lt. epsilon if n .gt. max(K, L) 297- 1 (x^n)/(1+x^n) = 0 for x = 0; for 0 .lt. x .lt. 1, x^n converges to 0, hence the limit (x^n)/(1+x^n) = 0. For x = 1, (x^n)/(1+x^n) = 1/2, hence the limit function is not continuous at 1 (in fact, the limit = 1 for x .gt. 1) 9 On [0,1], x^n .leq. 1, hence (x^n)/n converges uniformly to 0 = f. (f_n)' = (nx^(n-1))/n = x^(n-1) which converges to 0 for x .lt. 1, but is equal to 1, hence converges to 1 for x = 1. f'=0, but note that only the left hand derivative is defined for f at 1. 13 If a .gt. 0, (sin(nx)/nx) .lt. 1/na (since x .gt. a), hence the integral is .lt. (pi - a)(1/(na)) which goes to zero with n going to infinity. If a = 0, consider the range of integration as [0, epsilon/2] unipn [epsilon/2, pi]. since sin(x) .lt. x for x .gt. 0, the (sin(nx))/(nx) .lt. 1, hence the integral ove [0, epsilon/2] is less than epsilon/2 We have already seen that the integral over [epsilon/2, pi] goes to 0 for fixed epsilon/2, hence the limit is less than any given epsilon, hence the limit is 0.

Sun Jan 18 1998