Functions : Limits

Suppose that f:[0,1] -> R and f(a) = lim{x -> a} f(x) for all a in [0, 1]. Prove that f(q) = 0 for all q in Q intersection[0,1] implies that f(x) = 0 for all x in [0, 1]. Is the statement still true if f:[0, 1] -> R and f(a) = lim{x->a} f(x) for all a in Q intersection[0, 1]? R denotes the set of Real numbers Q denotes the set of Rational numbers Please explain thouroughly how to come up with the solution. Thank You.

Limits of Functions

Evaluate the following limits using the epsilon - delta definition and the limit theorems: a) lim {x -> 0} (x^2 + cos x)/(2 - tan x) b) lim {x -> sqrt(pi)} ((pi - x^2)^(1/3))/(x + pi)

Limits of Functions

Evaluate the following limits using the epsilon - delta definition and the limit theorems. a) lim {x -> 0} sin x sin (1/x^2) b) lim {x -> Infinity} (x^3 + 1)/(x^3 cos(1/x) + x^2 - 1) Please also show how you came up with the answer.

Continuous Functions

Where is the function f(x) = (q^2 - 1)/q^2 if x = p/q meaning x is a rational in reduced form and f(x) = 1 when x is not a rational continuous in the interval (0,1)? Please also explain how you came up with the answer.

Continuous Functions

Suppose that f(x) satisfies the functional equation f(x + y) = f(x) + f(y) for all x,y in R (the real numbers). Prove that if f(x) is continuous that f(x) = cx where c is a constant. What can you say about f(x) if it is allowed to be discontinuous?