$\lim_{x \to 2} \frac{\cos{\left(\frac{\pi}{x}\right)}}{x-2}$ without using De L'Hospital $$\lim_{x \to 2} \frac{\cos{\left(\frac{\pi}{x}\right)}}{x-2}$$
This limit is supposed be found without L'Hospital's Rule, but I have not been able to get close to the answer using conjugates, squares, Pythagorean Identity, half angle formulas or Squeeze Theorem. For each attempt, I expected at least one of  the well-known limits $\lim_{x \to 0} \frac{\sin{x}}{x}$ or $\lim_{x \to 0} \frac{\cos{x} - 1}{x}$ to come into use. They frequently did in my attempts but did not go anywhere. I have also been experimented substitutions but in vain.
How can this be solved without using L'Hospital's Rule?
 A: By definition of derivative, if $f(x)=\cos\left(\frac\pi x\right)$, then$$\lim_{x\to2}\frac{\cos\left(\frac\pi x\right)}{x-2}=f'(2)=\frac\pi4.$$
A: As an alternative, without derivatives, we have that
$$ \frac{\cos{\left(\frac{\pi}{x}\right)}}{x-2}
= \frac{\frac{\pi}2-\frac{\pi}{x}}{x-2}\frac{\sin{\left(\frac{\pi}2-\frac{\pi}{x}\right)}}{\frac{\pi}2-\frac{\pi}{x}}
=\frac{\pi(x-2)}{2x(x-2)}\frac{\sin{\left(\frac{\pi}2-\frac{\pi}{x}\right)}}{\frac{\pi}2-\frac{\pi}{x}}=$$
$$=\frac{\pi}{2x}\frac{\sin{\left(\frac{\pi}2-\frac{\pi}{x}\right)}}{\frac{\pi}2-\frac{\pi}{x}}\to \frac{\pi}4\cdot 1 =\frac{\pi}4$$
A: This expression corresponds with the definition of the derivative. Recall that $$f'(x) = \lim\limits_{y\to x}\frac{f(y)-f(x)}{y-x}$$ Using the fact that $\cos\left(\frac{\pi}{2}\right) = 0$, this expression turns out to be the derivative of $\cos(\frac{\pi}{x})$ at $x = 2$.
EDIT: Sorry, used the wrong function.
A: With $\dfrac\pi x=\dfrac\pi2-t$,
$$\lim\limits_{x \to 2} \frac{\cos{\left(\dfrac{\pi}{x}\right)}}{x-2}=\lim\limits_{t \to 0}\frac{\pi-2t}4\frac{\sin{\left(t\right)}}{t}.$$
No need to say more.
