Limit $x \to \pi/2$ without L'Hospital I'm having trouble solving this limit without L'Hospital:
$$ \lim_{x\to \pi/2} {\cos x\over x-\pi/2} $$
Thanks for any help. I have no idea, how expand. 
 A: Let $t = \pi/2-x$. Note that as $x \to \pi/2$, we have $t \to 0$. Also, recall that $\cos(x) = \sin(\pi/2-x)$.
Hence, we get that
$$\lim_{x \to \pi/2} \dfrac{\cos(x)}{x-\pi/2} = \lim_{x \to \pi/2} \dfrac{\sin(\pi/2-x)}{x-\pi/2} = \lim_{x \to \pi/2} \dfrac{\sin(\pi/2-x)}{-\left(\pi/2 -x\right)} =\lim_{t \to 0} \dfrac{\sin(t)}{-t} = -1$$
A: Note that this limit is exactly the definition of the derivative of $\cos x$ at $x=\pi/2$. So even if you're not using L'Hospital's rule to reach $\cos'(\pi/2)$, evaluating that will be exactly the same.
A: Where the sine function in radians crosses the $x$-axis, its slope is always $1$ or $-1$.  The shape of the graph of the cosine function is the same as that of the sine function; it's simply shifted horizontally.  So where the cosine function crosses the axis at $\pi/2$, going downward, its slope is $-1$.  The line $y=x-\pi/2$ also crosses at that same point, with a slope of $1$.  Looking at that point under a microscope, the graph of the cosine function looks like a line crossing at that point with slope $-1$, i.e. it looks like $y=-(x-\pi/2)$.
So it's as if you're looking at $\dfrac{-(x-\pi/2)}{x-\pi/2}=-1$.
