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.

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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 $$\lim_{x \to \pi/2} \dfrac{\cos(x)}{x-\pi/2} = \lim_{x \to \pi/2} \dfrac{\sin(\pi/2-x)}{x-\pi/2} = \lim_{x \to \pi/2} \dfrac{\sin(\pi/2-x)}{-\left(\pi/2 -x\right)} =\lim_{t \to 0} \dfrac{\sin(t)}{-t} = -1$$

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Of course, this assumes OP is familiar with $\lim_{t\to0}(\sin t/t)$ –  Gerry Myerson Nov 29 '12 at 23:04
@GerryMyerson True. But I guess it is a reasonable assumption to make. –  user17762 Nov 29 '12 at 23:09
-1: You're just calculating the derivative in a very roundabout way. Far too complicated and devoid of any elegance. –  commenter Nov 30 '12 at 0:47
@commenter I am interested in your easy and elegant way for computing the derivative of $\cos(x)$. –  user17762 Nov 30 '12 at 1:45
@commenter "You're just calculating the derivative in a very roundabout way. Far too complicated and devoid of any elegance." I am asking you what is the direct way (not roundabout way) to compute it? –  user17762 Nov 30 '12 at 2:31

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.

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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 $\dfrac{-(x-\pi/2)}{x-\pi/2}=-1$.

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If I didn't already know quite a lot about limits and derivatives, I wouldn't have been persuaded by that argument. –  mrf Nov 29 '12 at 23:18
@mrf : I would think if you know enough about limits and derivatives to understand the question, then that's enough. –  Michael Hardy Nov 29 '12 at 23:20
I disagree. The OP's question is a typical exercise in a first chapter about limits, before derivatives or "slopes" have been introduced. Many, if not most textbooks show that $\lim_{t\to0} \sin t/t = 1$ very early on. (Since the limit is typically used to compute the derivative of sine.) –  mrf Nov 29 '12 at 23:28
Apparently I assumed if he's mentioning L'Hopital's rule then he knows that stuff. But I suppose if all that is simply what he passed on to us from his instructor, that's another matter. –  Michael Hardy Nov 29 '12 at 23:45