# How to solve this limit without L'Hoptial's $\lim_{\Delta x \to 0} \frac{\sin\left(\frac{\pi}{6}+\Delta x\right)-\frac{1}{2}}{\Delta x}$?

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?

• First of all, you can pull out the $\frac{\sqrt{3}}{2}$. Then, we get $\frac{\sqrt{3}}{2} \lim \limits_{\Delta x \to 0} \frac{\sin{\Delta x}}{\Delta x} = \frac{\sqrt{3}}{2} \cdot 1 = \frac{\sqrt{3}}{2}$. – layman Feb 18 '15 at 17:03
• Looks like a derivative to me... – David Mitra Feb 18 '15 at 17:03
• Surely at this point the fact that $\lim \limits_{t\to 0}\left(\frac{\sin (t)}t\right)=1$ is available. – Git Gud Feb 18 '15 at 17:03
• It is a derivative, but my friend hasn't learned derivatives yet. I'm trying to do this strictly algebraically, and without any advanced math. – nw. Feb 18 '15 at 17:04
• See this. – David Mitra Feb 18 '15 at 17:10

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 $\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$.