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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 $?

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    $\begingroup$ I think it's correct. $\endgroup$ – Nosrati Jul 15 '18 at 4:53
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Your answer is absolutely correct.

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