Problem: Determine the limit l for the given a, and prove that it is the limit by showing how to find a δ such that abs(f(x) - l)< ε for all x satisfying 0< abs( x - a) < δ. ** abs = absolutue value of.... Solution: f(x) = x(3-cos(x^2)) The limit l is zero. hence the product of some number times zero is zero. abs(f(x) - 0)< ε = abs(x(3-cos(x^2)). abs((3-cos(x^2))≤ 1, abs(x(3-cos(x^2))< abs(x) Thus for every number x, If abs(x) < ε Then abs(x(3-cos(x^2)), Which finishes the proof, concluding that the limit of f(x) near zero is zero.