Let $f(x) = ext{sin} \, x$ - AQA - A-Level Maths Pure - Question 17 - 2017 - Paper 1

Using the identity for sine of a sum, we have:

$\sin(A + B) = \sin A \cos B + \cos A \sin B$

Thus, we can express:

$\sin \left( \frac{\pi}{6} + h \right) = \sin \left( \frac{\pi}{6} \right) \cos h + \cos \left( \frac{\pi}{6} \right) \sin h$

This allows us to simplify:

$= \lim_{h \to 0} \frac{\sin \left( \frac{\pi}{6} \right) \cos h + \cos \left( \frac{\pi}{6} \right) \sin h - \sin \left( \frac{\pi}{6} \right)}{h}$

Rearranging further, we have:

$= \lim_{h \to 0} \frac{\sin \left( \frac{\pi}{6} \right) (\cos h - 1) + \cos \left( \frac{\pi}{6} \right) \sin h}{h}$

This can be separated into two limits:

$= \lim_{h \to 0} \left( \sin \left( \frac{\pi}{6} \right) \frac{\cos h - 1}{h} + \cos \left( \frac{\pi}{6} \right) \frac{\sin h}{h} \right)$

Using the known limits:

1. ( \lim_{h \to 0} \frac{\sin h}{h} = 1 )
2. ( \lim_{h \to 0} \frac{\cos h - 1}{h} = 0 )

Thus, we have:

$= \sin \left( \frac{\pi}{6} \right) \cdot 0 + \cos \left( \frac{\pi}{6} \right) \cdot 1$

Knowing that ( \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} ) and ( \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} ), we get:

$= \frac{\sqrt{3}}{2}$

Therefore, the exact value is:

ight) = \frac{\sqrt{3}}{2}$$