I'm trying to help out a friend with Calc 1 and am struggling to find this limit without using l'hopital's or the small angle approximation.
$$\lim_{\Delta x \to 0} \frac{\sin\left(\frac{\pi}{6}+\Delta x\right)-\frac{1}{2}}{\Delta x}$$
Which I can reduce to
$$\lim_{\Delta x \to 0} \frac{\sqrt{3}\cdot\sin{\Delta x}}{2\cdot\Delta x}$$
Which is where I'm stuck. How can I simplify this further without the small angle approx or a taylor series expansion? Is there a way to do it with just trig identities?
Note that $\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}$, so we get $\frac{\mathrm{d}}{\mathrm{d}x}\left(\sin(x)\right)$ evaluated at $x=\frac{\pi}{6}$. This is just $\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$.
I think this is the intended method for the question, otherwise I don't think they would have chosen something that looks so similar to a derivative.