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I have been asked to find:

$\lim_{x\to 0} \, \cos \left(\frac{\pi -\pi \cos ^2(x)}{x^2}\right)$$=-1$

Without using l'Hôpital's rule. But I have no idea how to due it, could someone show a step by step process all the way to the answer?

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2 Answers 2

up vote 1 down vote accepted

Notice that:

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

As you know, the limit:

$$\lim_{x\to 0} \, \left(\frac{\sin x}{x}\right)=1$$

Therefore, our answer is:

$$\pi*1 = \pi$$

And

$$\cos(\pi) = -1$$

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Hint: We have $$\frac{\pi -\pi\cos^2 x}{x^2}=\frac{\pi\sin^2 x}{x^2}=\pi \left(\frac{\sin x}{x}\right)^2.$$

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I do not understand why (sinx)/x has limit of one as x approaches zero because sin[0]=1 and when the denominator approaches zero this number should be infinite large what am i missing? –  ALEXANDER Jul 30 at 2:32
    
The fact that $\lim_{x\to 0}\frac{\sin x}{x}=1$ is proved in calculus courses, usually geometrically, when one needs the derivative of $\sin x$, for this limit is a critical part of the proof. Note that $\sin 0=0$. If you look back to the differentiation of trigonometric functions part of your book, you will see this limit prominently discussed. –  André Nicolas Jul 30 at 2:50
    
Unfortunately my book only refers to a proof by the squeeze theorem, but It does not show the proof, would you be able to show me or should I pose a new question? –  ALEXANDER Jul 30 at 2:59
    
It has been done several times on MSE. Yes, it is a squeezing argument. The proof is in a great many places. If you ask the question on MSE, it will probably be quickly closed as a duplicate, but you will have at least one reference. I cannot show you in comments, it really requires a picture. –  André Nicolas Jul 30 at 3:19
    
Okey thank you! –  ALEXANDER Jul 30 at 5:04

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