For what value of the constant b is the function f continuous on (-infinity,infinity)

Question

f(x) =

x^6 cos(1/x) if x≠0

b if x=0

Answer 1

Hey Cassie,

I agree with Christopher. Essentially in calculus speak, you want to determine the limit of the function. If it exists, that is b, i.e.,

b = lim_x→0 x⁶cos(1/x)

There is a theorem we use in calculus called either the squeeze or sandwich theorem.

It says that if you have three functions f(x), g(x), and h(x), and it is so that g(x)≤f(x)≤h(x) for all x in your domain besides possibly at some point a in the domain, it is so that if

lim_x→a g(x) = lim_x→a h(x) = b,

then

lim_x→a f(x) = b.

The rest follows from Christopher’s solution...

We know the cosine function is trapped by -1 and 1. Here we manipulate the argument of the cosine function from x to 1/x but recall that will not change the amplitude of the function; so

-1≤cos(1/x)≤1.

Multiply this compound inequality by x⁶ yielding

-x⁶ ≤ x⁶cos(1/x) ≤ x^6.

Note,

lim_x→0 -x⁶ = lim_x→0 x⁶ = 0.

So by the squeeze or sandwich theorem

b = lim_x→0 x⁶cos(1/x) = 0.

Answer 2

Cassie:

Use the fact that for all x -1 ≤ cos(x) ≤ 1, or for this problem -1 ≤ cos(1/x) ≤ 1

Then -x⁶ ≤ x⁶*cos(1/x) ≤ x⁶

When x = 0, -x⁶ = 0 and x⁶ = 0 so that 0 ≤x⁶*cos(1/x)≤0

Therefore, choose f(x) = 0 when x =0 for continuity