Calculating the limit of a trigonometric function Calculate $$\lim_{x \to \pi /2 }{\frac{\cos x}{x-\frac{\pi}{2}}}$$
by relating it to a value of $(\cos x)'
$
My thoughts are to manipulate the limit algebraically and then just solve it. But how would this address the latter part of the question? How does this relate to -sinx?    
 A: Set $y = x-\pi/2$, such that your functions becomes $f(y) = \frac{\sin (y+\pi/2)}{y} = \frac{\sin (y)}{y}$, with the limit point $\pi/2$ becoming zero. The limit $\lim_{y \rightarrow 0} \frac{\sin (y)}{y}$ is zero, which can be seen e.g. from Taylor expansion of $sin (y)$.
A: First, we note that $\cos \pi/2=0$.  The derivative $f'(x_0)$ of a function $f$ at the point $x_0$ is given by
$$f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0} \tag 1$$
Now, we have
$$\lim_{x\to \pi/2}\frac{\cos (x)}{x-\pi/2}=\lim_{x\to \pi/2}\frac{\cos (x)-\cos (\pi/2)}{x-\pi/2} \tag 2$$
Comparing $(1)$ and $(2)$, we see that if $f(x)=\cos x$ and $x_0=\pi/2$, then 
$$\bbox[5px,border:2px solid #C0A000]{\frac{d\,\cos x}{dx}=\lim_{x\to \pi/2}\frac{\cos (x)}{x-\pi/2}=-\sin (\pi/2)=-1}$$
A: L'Hôpital's Rule relates the limit to the derivative of $\cos x$, since substitution of $\pi /2$ in the original limit yields $0/0$.  
Taking the derivative of each of the numerator and the denominator gives $-\sin x$. We can simply plug in $\pi /2$ and obtain $-1$.  
$$\lim_{x \to \pi /2 }{\frac{\cos x}{x-\frac{\pi}{2}}}=\lim_{x \to \pi /2 }{\frac{d/dx(\cos x)}{d/dx(x-\frac{\pi}{2})}} =\lim_{x \to \pi /2 }(-\sin x)=\lim_{x \to \pi /2 }(-\sin( \frac{\pi}{2}) )= -1$$ 
