$\displaystyle\lim\limits_{x\to 0} \frac{\sin\left(\frac{\pi}{2}+x\right)-1}{x}=$

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Substituting the value $x=0$ results in the indeterminate form. We can use L'Hospital's rule.

$$\lim\limits_{x\to 0} \frac{\sin\left(\frac{\pi}{2}+x\right)-1}{x}= \lim\limits_{x\to 0} \frac{\cos(\frac{\pi}{2}+x)}{1}$$ $$= \cos\frac{\pi}{2} = \boxed{0}$$

Comparing it with the limit definition of a derivative:

$$\lim \limits_{h\to 0} \frac{f(x+h)-f(x)}{h}$$ $$\lim \limits_{h\to 0} \frac{\sin(\frac{\pi}{2}+h)-\sin\left(\frac{\pi}{2}\right)}{h}$$ $$f(x)=\sin x \hskip{2em} x=\frac{\pi}{2}$$ $$f'(x)=\cos x$$ $$f'\left(\frac{\pi}{2}\right)=\cos \frac{\pi}{2} = \boxed{0}$$