This is a duplicate of this post here, but I'm new to Stack Exchange so I'm not able to comment on that post yet (and sorry in advance if this question isn't appropriate to the platform).
I was reading through Spivak's Calculus (4th ed.) and came across Problem 5.3.i asking to "determine the limit l for the given a, and prove that it is the limit by showing how to find a $\delta$ such that $|f(x)-l|< \epsilon$ for all x satisfying $0<|x-a|<\delta$: $f(x)=x[3-cos(x^2)], a=0$. Not sure if my solution is correct, but here's what I have:
I first found the limit by just substituting $0$ into the different parts of the function ($3x$, $x$, $cos(x^2)$) and multiplied and added the limits to get $l=0$.
Then substituting values into $|f(x)-l| < \epsilon$, we obtain $|x[3-cos(x^2)]| < \epsilon$. By properties of absolute values, factor and divide to get $|x|<\frac{\epsilon}{|3-cos(x^2)|}$, and we get desired form of $|x-a|<\delta$ (since $a=0$).
So let $\delta = \frac{\epsilon}{|3-cos(x^2)|}$, but since the minimum value of $\delta$ is when the denominator is maximized (so when $cos(x^2) = -1$, we can simplify to $\delta = \frac{\epsilon}{4}$, completing the proof.
Is this the proper way to approach the problem, and is the final solution correct? Also, is simplification acceptable in the final step, or does it make the answer less precise?
