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$$\lim_{x \to 2} \frac{\cos{\left(\frac{\pi}{x}\right)}}{x-2}$$

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?

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  • $\begingroup$ Someone did downvote all answers, but s/he forgot to do downvote the question $;)$. $\endgroup$ – Nosrati Sep 7 '18 at 5:15
  • $\begingroup$ @Nosrati: I am OP and I did not mark a single answer down, so please be more hesitant to pass judgement $\endgroup$ – hatinacat2000 Sep 7 '18 at 17:20
  • $\begingroup$ I didn't say you are downwoter! I said that for fun dear. Anyway I did delete my answer. $:)$ @hanti $\endgroup$ – Nosrati Sep 7 '18 at 20:04
  • $\begingroup$ @hatinacat2000 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details HERE $\endgroup$ – user Oct 23 '18 at 21:14
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By definition of derivative, if $f(x)=\cos\left(\frac\pi x\right)$, then$$\lim_{x\to2}\frac{\cos\left(\frac\pi x\right)}{x-2}=f'(2)=\frac\pi4.$$

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    $\begingroup$ intuition is a gift+ $\endgroup$ – Mikasa Sep 6 '18 at 18:28
  • $\begingroup$ Answer should be $\pi/4$. $\endgroup$ – A B Sep 6 '18 at 18:31
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    $\begingroup$ @StammeringMathematician I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Sep 6 '18 at 18:32
  • $\begingroup$ IMO, using this trick is a little cheating. There is very little difference between $(f(x)-f(x_0))/(x-x_0)\to f'(x_0)$ and $(f(x)-f(x_0))/(x-x_0)\to f'(x_0)/1$. $\endgroup$ – Yves Daoust Sep 6 '18 at 19:05
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As an alternative, without derivatives, we have that

$$ \frac{\cos{\left(\frac{\pi}{x}\right)}}{x-2} = \frac{\frac{\pi}2-\frac{\pi}{x}}{x-2}\frac{\sin{\left(\frac{\pi}2-\frac{\pi}{x}\right)}}{\frac{\pi}2-\frac{\pi}{x}} =\frac{\pi(x-2)}{2x(x-2)}\frac{\sin{\left(\frac{\pi}2-\frac{\pi}{x}\right)}}{\frac{\pi}2-\frac{\pi}{x}}=$$

$$=\frac{\pi}{2x}\frac{\sin{\left(\frac{\pi}2-\frac{\pi}{x}\right)}}{\frac{\pi}2-\frac{\pi}{x}}\to \frac{\pi}4\cdot 1 =\frac{\pi}4$$

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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.

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  • $\begingroup$ Can you please show your work? I have tried your substitution but am not able to get this result. Also, how did you come up with this substitution? $\endgroup$ – hatinacat2000 Sep 7 '18 at 5:03
  • $\begingroup$ @hatinacat2000: this is elementary algebra. And seeing the shape of the expression ($0/0$ with a trigonometric function), I understood that it could be turned to $\sin t/t$ by an appropriate transform. $\endgroup$ – Yves Daoust Sep 7 '18 at 7:12
  • $\begingroup$ I can get $\frac {sin {t}}{t} $, but not $(\pi - 2t)/4$ $\endgroup$ – hatinacat2000 Sep 7 '18 at 17:19
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    $\begingroup$ Edit: success after more substitutions $\endgroup$ – hatinacat2000 Sep 7 '18 at 17:50
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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.

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