back to limits.

[africanviolin]

Prove.

lim x--> 0 x^2cos(1/x^2)=0

so the lim x-->0 x^2 = 0 and anything multiplied by zero is zero. is that it? b/c the cos lim is undefined @ 0 and oscillates b/w a neg and pos when x-values approach 0 from either side. is there a squeeze thm opportunity i'm missing? i don't get it :(

ps i'm studying for the final on thurs...hopefully i

Comments

[joshua_green]

That argument is almost right.  If limx→0 cos(1/x²) existed, then that would be a valid argument (since the limit of a product is the product of the limits when the limits in the product exist).  In this case, you can't really do that.  However, since cos is bounded, you shouldn't have too much difficulty finding δ's for your ε's.

[joshua_green]

Alternatively, a squeeze theorem could be applied, but I'll leave it to you to find the appropriate bounding functions.  (Hint: ∀ θ ∈ R, |cos(θ)| ≤ 1.)

[hawkington]

More generally, suppose f is bounded and g goes to zero as x->a. Can you show that the product fg goes to zero as x->a?