Prove that $\sin(ax+b)$ is continuous using $|\sin x| < |x|$
I would like to ask for verification of my proof. Essentially to obtain a proof we need to show that: $$ \lim_{x\to x_0}\sin(ax+b) = \sin(ax_0 + b) $$ or: $$ |x-x_0| < \delta_\epsilon \implies |\sin(ax+b) - \sin(ax_0+b)| < \epsilon $$
I've started with rewriting the RHS of the implication: $$ \begin{align} |\sin(ax+b) - \sin(ax_0+b)| &= \left|2\sin\left(\frac{ax+b - ax_0 - b}{2}\right)\cos\left(\frac{ax + b + ax_0 + b}{2}\right)\right| \\ &= \left|2\sin\left(\frac{ax+b - ax_0 - b}{2}\right)\right|\left|\cos\left(\frac{ax + b + ax_0 + b}{2}\right)\right| \\ &= \left|2\sin\left(\frac{ax- ax_0}{2}\right)\right|\underbrace{\left|\cos\left(\frac{ax + ax_0 + 2b}{2}\right)\right|}_{\le 1} \\ &\le \left|2\sin\left(\frac{ax- ax_0}{2}\right)\right| \\ &\le \left|2\left(\frac{ax- ax_0}{2}\right)\right| \\ &= |ax-ax_0| \\ &= |a||x-x_0| \end{align} $$
Now chose $\delta = {\epsilon \over |a|}$, then: $$ \forall x:|x-x_0| < \delta \implies |\sin(ax + b) - \sin(ax_0 + b)| < \epsilon $$
I have some doubts regarding the last statement. Could you please confirm my reasoning is fine or propose an improvement/correct-proof in case the above makes no sense.
The proof is correct except for a minor glitch. If $a=0$ you cannot set $\delta=\epsilon/|a|$, but that's easily fixed: in this case set $\delta=1$ (or your preferred positive number).
You have indeed proved that, for $a\ne0$, if $|x-x_0|<\delta=\varepsilon/|a|$ one has $$ \lvert\sin(ax+b)-\sin(ax_0+b)\rvert\le |a||x-x_0|<|a|\frac{\epsilon}{|a|}=\epsilon $$ and therefore $$ \lvert\sin(ax+b)-\sin(ax_0+b)\rvert<\epsilon $$