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I need to calculate the following limit without using L'Hospital's rule:

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

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  • $\begingroup$ I see the problem statement, but missing a question in this post. $\endgroup$ – A.Γ. Feb 20 at 13:48
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Let $t=2-x$. Then,

$$\lim_{x \to 2} \frac{\cos(\pi/x)}{2-x}=\lim_{t \to 0} \frac{\cos\frac{\pi}{2-t}}{t} =\lim_{t \to 0} \frac{\sin\frac{t\pi}{2(t-2)}}{t} =\lim_{t \to 0} \frac{\sin\frac{t\pi}{2(t-2)}}{\frac{t\pi}{2(t-2)}}\lim_{t \to 0}\frac{\pi}{2(t-2)}=-\frac\pi4$$

where $\lim_{u \to 0}\frac{\sin u}{u} =1$ is used.

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$x=2+y$, then $$L=\lim_{x \rightarrow 2} \frac{\cos(\pi/x)}{2-x} \lim_{ \rightarrow 0} \frac{\cos(\pi/(2+y))}{-y}= \lim_{y \rightarrow 0} \frac{\cos(\pi/2(1-y/2))}{-y}=\lim_{y \rightarrow 0} \frac{\cos(\pi/2-\pi y/4)}{-y}=\lim_{x \rightarrow 0} \frac{\sin (\pi y/4)}{-y}=-\frac{\pi}{4}$$. In above we have used $\frac{\pi}{2+y}=\frac{\pi}{2}(1+y/2)^{-1} \approx\frac{\pi}{2}(1-y/2)$ (Binomial approx.)

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Hint

Use $$\cos\dfrac\pi x=\sin\left(\dfrac\pi2-\dfrac\pi x\right)=\sin\dfrac{\pi(x-2)}{2x}$$

to find $$-\lim_{x\to2}\dfrac{\sin\dfrac{\pi(x-2)}{2x}}{\dfrac{\pi(x-2)}{2x}}\cdot\lim_{x\to2}\dfrac\pi{2x}=?$$

and Why the limit of $\frac{\sin(x)}{x}$ as $x$ approaches 0 is 1?

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    $\begingroup$ I understand how you transformed cos into sin. I did not understand how you separated the two limits. $\endgroup$ – Ciro Nogueirão Shia Feb 20 at 14:34
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    $\begingroup$ @Ciro, it's legitimate in product if at least one of the limits is non zero and finite $\endgroup$ – lab bhattacharjee Feb 20 at 14:40

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