# Continuity of piecewise defined trig functions

## Homework Statement

Define functions f and g on [-1,1] by

f(x) = xcos(1/x) if x≠0 and 0 if x = 0

g(x)= cos(1/x) if x≠0 and 0 if x = 0

(These are piecewise defined. I don't know how to type them in here.)

Prove that f is continuous at 0 and that g is not continuous at 0. Explain why these functions are continuous at every other point in [-1,1].

## The Attempt at a Solution

Using the definition of continuity,

Suppose g is continuous at x=0.

Since g is continuous for all x in [-1,1], then for all ε>0, there exists a δ > 0 such that |g(x)-g(c)| < ε for all x in [-1,1] that satisfy |x-c| < δ

Let ε=1/2
Let c = 0
Choose δ=ε

Then |g(x)-g(c)| = |cos(1/x)| and since -1 ≤ cos(1/x) ≤ 1 for all x in [-1,1], then |cos(1/x)|≤|2x|<2δ=2/2=1

Then if x=3/$\pi$, |cos(1/x)| = |1/2| < 1/2 which 1/2 cannot be less than itself. So there exists a ε where the continuity definition fails at c=0 and thus g(x) is not continuous on [-1,1].

Is this correct for showing g(x) is not continuous at x=0?

I am unsure how to show f(x) is continuous at x=0 since it is a product of two functions where one of them is not continuous at 0.

I have a theorem that will help me with showing that trig functions are continuous if they are defined on their domain.

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jbunniii
Homework Helper
Gold Member
I didn't understand your argument for proving that g is not continuous at 0. I don't think it's right. Can you show that in any neighborhood around x = 0, g(x) takes on the values +1 and -1 for infinitely many x? Can you use that fact to show that g is not continuous at 0?

To show that f is continuous at 0, try sandwiching it between two continuous functions which both have the same value at x = 0.

HallsofIvy