Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 5 down vote accepted

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

share|cite|improve this answer
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.

share|cite|improve this answer

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.