Consider the function f(x) = cos (1/x), x are real numbers and not equal to zero.

Question

a) For positive integer n, find the values of f(x) when x = [1/(nπ +π/2)] , (1/2nπ) and [1/(2n+1)π].

b) Do the limits lim x→0 cos(1/x) and lim x→0 [cos (1/x) / cos (1/x) ] exist? Why?

c) Consider the functions g(x) = x f(x). How should g(x) be defined at x=0 so that it is continuous?

Answer 1

(a) When x=1/(nπ+π/2), 1/x=nπ+π/2 and cos(1/x)=0. When x=1/(2nπ), cos(1/x)=1. When x=1/((2n+1)π), cos(1/x)=-1

(b) Limit as x→0 of cos(1/x): 1/x cannot be defined, so cos(1/x) is undefined, value fluctuates between -1 and +1. But the same limit for cos(1/x)/cos(1/x) does exist and equals 1, because the denominator and numerator have the same value. The expression is undefined at x=0, but we are concerned only about x<0 or x>0.

(c) g(x)=xf(x). When x→0 g(x)→0 because f(x) ranges between -1 and +1 (has a finite value). At x=0 therefore, the product of x and f(x) must be zero, so g(x)=0.