\( \lim _ { h \rightarrow 0 } \frac { \sin ( x + h ) - \sin x } { h } \rightarrow \)

To evaluate the limit \[ \lim_{h \rightarrow 0} \frac{\sin(x + h) - \sin x}{h}, \] we can use the definition of the derivative of the sine function. The expression inside the limit is the difference quotient for the derivative of \(\sin x\) at the point \(x\). Using the sine addition formula, we have: \[ \sin(x + h) = \sin x \cos h + \cos x \sin h. \] Substituting this into the limit gives: \[ \lim_{h \rightarrow 0} \frac{\sin(x + h) - \sin x}{h} = \lim_{h \rightarrow 0} \frac{(\sin x \cos h + \cos x \sin h) - \sin x}{h}. \] This simplifies to: \[ \lim_{h \rightarrow 0} \frac{\sin x \cos h - \sin x + \cos x \sin h}{h} = \lim_{h \rightarrow 0} \frac{\sin x (\cos h - 1) + \cos x \sin h}{h}. \] Now we can separate the limit into two parts: \[ \lim_{h \rightarrow 0} \left( \frac{\sin x (\cos h - 1)}{h} + \frac{\cos x \sin h}{h} \right). \] 1. For the first term, as \(h \rightarrow 0\), \(\cos h \rightarrow 1\), so \(\cos h - 1 \rightarrow 0\). We know that \(\cos h - 1 \approx -\frac{h^2}{2}\) for small \(h\), thus: \[ \frac{\sin x (\cos h - 1)}{h} \approx \frac{\sin x \left(-\frac{h^2}{2}\right)}{h} = -\frac{\sin x}{2} h \rightarrow 0 \text{ as } h \rightarrow 0. \] 2. For the second term, we use the fact that \(\sin h \approx h\) as \(h \rightarrow 0\): \[ \frac{\cos x \sin h}{h} \approx \frac{\cos x \cdot h}{h} = \cos x. \] Putting it all together, we find: \[ \lim_{h \rightarrow 0} \left( 0 + \cos x \right) = \cos x. \] Thus, the limit is: \[ \lim_{h \rightarrow 0} \frac{\sin(x + h) - \sin x}{h} = \cos x. \] Therefore, the final answer is: \[ \boxed{\cos x}. \]