When reading Spivak's calculus book, I stumbled upon this limit: $$ \lim_{x \to 0} x \ [3-\cos(x^2)] = 0 $$
Proof:
We have that $$0 < |x| < \delta $$ Also $$ |x \ (3-\cos(x^2))| = |x| \ |3-\cos(x^2)| < \epsilon$$ Since, $$ |3-\cos(x^2)| \le 4$$ we can write that: $$ |x| \ |3-\cos(x^2)| < 4|x| < \epsilon$$ $$\therefore |x| < \epsilon \ / \ 4$$ $$\therefore \delta = \epsilon \ / \ 4$$
By letting $\delta = \epsilon \ / \ 4$, we get that $|x \ (3-\cos(x^2))|<\epsilon$ if $0 < |x| < \delta $.
Thus, $ \lim_{x \to 0} x \ [3-\cos(x^2)] = 0 $.
Since I don't have the book with answers, I can't verify my solution (the book is on its way).
Is there anything that I missed here? Maybe there is another way, more concise perhaps, of finding a $\delta $?
Your answer is absolutely correct.