Having some trouble with this question and its really bugging me!

$$ \lim_{x\to 1} \frac{\cos (\pi x/2))}{1-x} $$

I know I can solve this limit using L'Hôpital's Rule but I'm asked to solve the limit without using L'Hôpital's Rule (or Talyor series).

Many thanks.

  • 1
    $\begingroup$ What can you use? $\endgroup$ – José Carlos Santos Apr 23 '18 at 15:56
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    $\begingroup$ As a matter of definition, this is the derivative of $\cos(\pi x / 2)$ at $1$. $\endgroup$ – user296602 Apr 23 '18 at 15:57
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    $\begingroup$ Try with $$1-x=h$$ $\endgroup$ – lab bhattacharjee Apr 23 '18 at 15:58
  • $\begingroup$ Okay thank you I will try doing so. $\endgroup$ – B.Hull Apr 23 '18 at 16:03

As almost always, I like limits where the variable goes to zero.

So, let $x = 1+y$. Then

$\begin{array}\\ \lim_{x\to 1} \frac{\cos (\pi x/2))}{1-x} &=\lim_{y\to 0} \frac{\cos (\pi (1+y)/2))}{1-(1+y)}\\ &=\lim_{y\to 0} \frac{\cos(\pi/2)\cos(\pi y/2)-\sin(\pi/2)\sin(\pi y/2)}{-y}\\ &=\lim_{y\to 0} \frac{-\sin(\pi y/2)}{-y}\\ &=\lim_{y\to 0} \frac{\sin(\pi y/2)}{y}\\ \end{array} $

If you know that $\lim_{y\to 0} \frac{\sin(y)}{y} =1$, then you are done (with a little algebra, because you can write $\frac{\sin(\pi y/2)}{y} =\frac{\sin(\pi y/2)}{\pi y/2}(\pi/2) \to \pi/2 $).


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