# Finding $\delta$ for $\lim_{x \to 0} x \ [3-\cos(x^2)] = 0$

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 $\delta$?

• I think it's correct. – Nosrati Jul 15 '18 at 4:53