How to find $\lim_{x \to \frac{\pi}{6}}\frac{\sin(x)-\frac{1}{2}}{x-\frac{\pi}{6}}$ without using L'Hospital's Rule? I have to find $$\lim_{x \to \frac{\pi}{6}}\frac{\sin(x)-\frac{1}{2}}{x-\frac{\pi}{6}}$$
We are not allowed to use L'Hospital's rule, any suggestions would be beneficial!
I have tried multiplying by the conjugate (both the numerator and the denominator).
I have tried using trig substitution. 
 A: HINT: Use the definition for derivatives on the function $f(x) = \sin x$ and use the fact that $\sin(\frac{\pi}{6}) = \frac 12$
A: Let $x'=x-\pi/6$.  Then
$$\lim_{x' \to 0} \frac{\sin(x'+\pi/6)-\frac{1}{2}}{x'}$$
Then apply the addition formula for sine, and it will reduce to
$$\lim_{x' \to 0} \frac{\sin(x')\cos(\pi/6)+\cos(x')sin(\pi/6)-\frac{1}{2}}{x'}$$
$$\lim_{x' \to 0} \frac{\sin(x')\cos(\pi/6)}{x'}+\frac{\cos(x')sin(\pi/6)-\frac{1}{2}}{x'}$$
$$=\frac{\sqrt{3}}{2}+0$$
A: Using the definition of the derivative of $\sin(x)$ at $x = \dfrac{\pi}{6}$, we get
\begin{align}
   \lim_{x \to \frac{\pi}{6}}\frac{\sin(x)-\frac{1}{2}}{x-\frac{\pi}{6}}
   &=\lim_{x \to \frac{\pi}{6}}
     \frac{\sin(x)-\sin(\frac{\pi}{6})}{x-\frac{\pi}{6}} \\
   &= \sin'(\frac{\pi}{6}) \\
   &= \cos(\frac{\pi}{6}) \\
   &= \dfrac{\sqrt 3}{2}
\end{align}
A: Just another way to do it.
Considering $$A=\frac{\sin(x)-\frac{1}{2}}{x-\frac{\pi}{6}}$$ make life simpler using $x=y+\frac{\pi}{6}$ and expand $\sin(y+\frac{\pi}{6})$. 
This makes $$A=\frac {\sqrt{3} \sin (y)+\cos (y)-1}{2y}$$ Now, let us Taylor expansions around $y=0$; this gives $$\sqrt{3} \sin (y)+\cos (y)-1=\sqrt{3} y-\frac{y^2}{2}+O\left(y^3\right)$$ and so $$A=\frac{\sqrt{3}}{2}-\frac{y}{4}+O\left(y^2\right)$$ which shows the limit and how it is approached when $y\to 0$.
Edit
If you add the next term of the Taylor expansion, you would get $$A=\frac{\sqrt{3}}{2}-\frac{y}{4}-\frac{y^2}{4 \sqrt{3}}+O\left(y^3\right)$$ You could me amazed to see how close are the original function and the above approximation for the range $-1\leq y \leq 1$. This means that, if, at any time, you need to find $x$ such that $A(x)=a$,$(a>0)$ you would a nice approximation solving for $y$ the quadratic$$a=\frac{\sqrt{3}}{2}-\frac{y}{4}-\frac{y^2}{4 \sqrt{3}}$$ the solution of which being $$y=\frac{ \sqrt{27-16 \sqrt{3} a}-\sqrt 3}{2 }$$ For illustration purposes, let us use $a=\frac 12$; this would give an approximate solution $y\approx 0.9467$ while the "exact" solution would be $y\approx 0.9975$ which is not too bad.
A: Hint:
Think of the rate of variation $\;\dfrac{\Delta f}{\Delta x}=\dfrac{f(x')-f(x)}{x'-x}$.
A: Using the derivative of sin (x) at $\pi/6$
$$\lim_{x \to \frac{\pi}{6}}\frac{\sin(x)-\frac{1}{2}}{x-\frac{\pi}{6}}=\lim_{x \to \frac{\pi}{6}}\frac{\sin(x)-\sin(\frac{\pi}{6})}{x-\frac{\pi}{6}}=\sin'(\frac{\pi}{6})
=\cos(\frac{\pi}{6})={\sqrt{3}}/2$$
