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.
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 $).