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So I ran into this problem:

$\lim _{x\to \pi }(\frac{\cos\left(\frac{x}{2}\right)}{\pi ^2-x^2})$

We haven't studied L'Hôpital's rule in my class yet, we basically need to transform the expression using trigonometric identities and use known limits like $\lim_{x \to 0} \frac{sinx}{x} = 1$ to calculate in case we get an indefinite form.

I couldn't find anything I could do to calculate this limit :(

Some help?

Thanks.

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$$\ \lim_{x\to\pi}\frac{\cos(\frac{x}{2})}{\pi^2-x^2}=\lim_{x\to\pi}\frac{\sin(\frac{\pi}{2}-\frac{x}{2})}{(\pi+x)(\pi-x)}$$

And now let $\pi-x=t, t\to0$

So you get: $$\lim_{t\to0}\frac{\sin(\frac{t}{2})}{t}\cdot\frac{1}{2\pi-t}=\frac{1}{2}\cdot\frac{1}{2\pi}=\frac{1}{4\pi}$$

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Hint: Notice that $$\frac{\cos x/2}{\pi^2 - x^2} = \frac{1}{-2} .\frac{1}{x + \pi} . \frac{\cos x/2}{x/2 - \pi/2} = \frac{1}{2}\frac{1}{x + \pi} . \frac{\sin\left(\frac{x}{2} - \frac{\pi}{2} \right)}{\frac{x}{2} - \frac{\pi}{2}}$$

Now you can evaluate the limit of that expression of $x \to \pi$.

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If you want to use known limits you should have something that goes to $0$. So substitute $x = t + \pi$ to get $$\lim_{t \to 0}-\frac{\cos\left(\frac t2 + \frac \pi2\right)}{t(t + 2\pi)} = \lim_{t \to 0}-\frac{-\sin\frac t2}{t(t + 2\pi)} = \frac12\lim_{t \to 0}\frac{\sin\frac t2}{\frac t2}\times\lim_{t \to 0}\frac 1{t + 2\pi} = \frac 1{4\pi}$$

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