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I have to evaluate this limit without using L'Hopital. Could you help me

$$\lim_{x \to 0} {1-\sqrt{\cos(x)}\over x^2}$$

I already rationalized it:

$$\lim_{x \to 0} \left({1-\sqrt{\cos(x)}\over x^2}\right) \left({1+\sqrt{\cos(x)}\over 1+\sqrt{\cos(x)}}\right)$$

And I got:

$$\lim_{x \to 0} \left({1-\cos(x)\over x^2(1+\sqrt{\cos(x)})}\right)$$

What should I do next?

Cameron Williams
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Manuel
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  • Of the three factors there, one of them becomes nice as $x\to0$. Can you use that to simplify your limit? – Semiclassical Sep 07 '14 at 02:36
  • Yes, but the result still being 0/0 – Manuel Sep 07 '14 at 02:38
  • Sure. I just mean that your rationalizing it means that the $1+\sqrt{\cos(x)}$ term isn't something you need to worry about anymore. That still leaves the remainder to consider; an easy way to find the rest of the limit is by expanding the denominator in a Taylor series. (That assumes you know how to do so, of course.) – Semiclassical Sep 07 '14 at 02:42
  • You get $1+\sqrt{\cos(x)} \to 2$ as $x \to 0$. So it remains to compute $\lim_{x \to 0} \frac{1-\cos(x)}{x^2}$. This is a standard limit and can be done with purely geometric methods. – Ian Sep 07 '14 at 02:50
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    Use $1-\cos x=2\sin^{2}(x/2)$ – Paramanand Singh Sep 07 '14 at 02:51

1 Answers1

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Note that $$ \frac{1-\cos(x)}{x^2(1+\sqrt{\cos x})}=\frac{1-\cos^2 x}{x^2(1+\sqrt{\cos x})(1+\cos x)}=\left(\frac{\sin x}{x}\right)^2\frac{1}{(1+\sqrt{\cos x})(1+\cos x)} $$ then you can use $\lim_{x\to 0}\sin(x)/x=1$. Arguments for this latter fact can be seen here. Some of them do not employ L'Hopital's Rule.

Community
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Kim Jong Un
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  • Given that the poster specifically indicates that L'Hopital's rule isn't to be used, I think you need to show why $\sin(x)/x\to 1$ as $x\to0$ is valid. – Semiclassical Sep 07 '14 at 03:29
  • @Kim Jong Un, excuse me, Where did you get the $(1+cos(x))$ ? – Manuel Sep 07 '14 at 03:30
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    @ Semiclassical: I added a link. @ Manuel: I multiplied both the numerator and denominator by $1+\cos x$. The numerator then becomes $1^2-\cos^2x=\sin^2x$. – Kim Jong Un Sep 07 '14 at 03:32
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    @AntonioVargas: Fair point, especially considering (as noted in one of the comments in the question linked by KimJongUn) that $\sin(x)/x\to 0$ as $x\to 0$ is equivalent to the first derivative of $\sin x$ being 1 at zero. So using L'Hopital would be circular anyways... – Semiclassical Sep 07 '14 at 03:34