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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.

amWhy
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iymz
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3 Answers3

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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. –  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 00:47
  • @commenter I am interested in your easy and elegant way for computing the derivative of $\cos(x)$. –  Nov 30 '12 at 01:45
  • Plug the definition into your favorite definition and say what you're doing. – commenter Nov 30 '12 at 01:51
  • @commenter I am interested in how you compute the derivative of $\cos(x)$ without making use of the "complicated and inelegant" fact that $$\lim_{t \to 0} \dfrac{\sin(t)}t = 1$$ –  Nov 30 '12 at 01:54
  • If you put things into quotation marks you should make sure that you quote properly. – commenter Nov 30 '12 at 01:58
  • @commenter So you do not know of an uncomplicated and elegant way of computing the derivative of $\cos(x)$. Interesting... –  Nov 30 '12 at 02:08
  • You are well aware that this is a non-sequitur. I'm out of this discussion. – commenter Nov 30 '12 at 02:11
  • @commenter Lol. A down-voter with no proper reason. –  Nov 30 '12 at 02:12
  • I did give you my reasons: a) you're computing the derivative b) you didn't say so. Whatever your interest in how I calculate this is or whether you consider it proper has nothing to do with it. Throwing lolz at me doesn't change this. – commenter Nov 30 '12 at 02:29
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    @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? –  Nov 30 '12 at 02:31
  • @MattN. First what you have is incorrect $\lim_{x \to \pi/2} \dfrac{\sin(\pi/2-x)}{\pi/2-x} \neq \left. \dfrac{d \sin(x)}{dx} \right \vert_{x = \pi/2}$. and my question is how do you evaluate the derivative? You need to make use of $\lim_{t \to 0} \dfrac{\sin(t)}t = 1$. And I assume in the last expression you meant evaluating the derivative at $x = 0$ and not $x = \pi/2$. –  Nov 30 '12 at 07:48
  • Duh. Ignore what I wrote above. I'll delete my comments since they add nothing. – Rudy the Reindeer Nov 30 '12 at 07:52
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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.

hmakholm left over Monica
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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 $\dfrac{-(x-\pi/2)}{x-\pi/2}=-1$.

Michael Hardy
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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