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I got problem with my limit problem. I don't know what I should to do next. There are two numbers that I can't find. Please help me to do this.

  1. $\displaystyle \lim_{x \to \pi/2} \frac{1-\cos 2x}{2\cos x}$
  2. $\displaystyle \lim_{x \to 0} \frac{1 -2\cos x + \cos 2x}{x^2}$
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first one is easy as denominator tends to $2$ and denominator tends to $0$. Clearly from LHS limit is $\infty$ and from RHS it is $-\infty$. Hence limit does not exist. For second one note that numerator is $2\cos x(\cos x - 1)$ and clearly $(\cos x - 1)/x^{2} \to -1/2, \cos x \to 1$ as $x \to 0$. – Paramanand Singh Aug 20 '15 at 5:11
up vote 3 down vote accepted

Presumably the idea is to use De L'Hôpital's Rule (see here - particularly Corollary 1).

I could help you further if you have trouble differentiating.

EDIT: Of course, $\cos \pi = -1$, so that the first limit does not exist (denominator diverges). My apologies.

EDIT 2: While I see others are making errors in the application of De L'Hôpital's Rule (HR from now), let me apply it to the second case:

We have that $1-2\cos 0 +\cos 2\cdot 0 = 1 -2 +1=0 = 0^2$, so that HR yields:

$$\lim_{x\to 0} \frac{1-2\cos x+\cos2x}{x^2} = \lim_{x\to0} \frac{2\sin x-2\sin2x}{2x}$$

We see that it is required to apply HR again, this time yielding:

$$\lim_{x\to0} \frac{2\sin x-2\sin2x}{2x} = \lim_{x\to0} \frac{2\cos x-4\cos 2x}{2}$$

This latter limit we can simply evaluate by continuity of the involved functions; it follows that the limit is $-1$. In summary, we conclude:

$$\lim_{x\to 0} \frac{1-2\cos x+\cos2x}{x^2} = -1$$

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BTW, $1-2\cos x+\cos2x= -x^2+7 x^4/12-31 x^6/360+O(x^8)$ – lhf Oct 9 '12 at 18:33
yes, its very helpful, thanks again – NPE Oct 11 '12 at 15:34

Implement the formula:

$1)$ $1=\cos^2 (x)+\sin^2 (x)$

$2)$ $\cos (2x)=\cos^2 (x)-\sin^2 (x)$

$\lim_{x\to\frac{\pi}{2}}\frac{1-\cos(2x)}{2\cos(x)}$=$\lim_{x\to\frac{\pi}{2}}\frac{1-(\cos^2(x)-\sin^2(x))}{2\cos(x)}$=$\lim_{x\to\frac{\pi}{2}}\frac{\cos^2 (x)+\sin^2 (x)-\cos^2(x)+\sin^2(x)}{2\cos(x)}$=$\lim_{x\to\frac{\pi}{2}}\frac{2\sin^2 (x)}{2\cos(x)}$=$\lim_{x\to\frac{\pi}{2}}\frac{\sin^2 (x)}{\cos(x)}=\lim_{x\to\frac{\pi}{2}}\frac{(\sin\frac{\pi}{2})^2}{\cos\frac{\pi}{2}}=\lim_{x\to\frac{\pi}{2}}\frac{1^2}{0}=\frac{1}{0}???$



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Umm, so your answer is ??? ? You can see that it is of the form $\frac{2}{0}$ without doing any work other than plugging in $\pi/2$. So, your answer uses multiple trig identities and many steps to arrive at a conclusion that could have been obtained immediately. – Graphth Oct 9 '12 at 22:41

To do the first limit, your first step is to "plug" in $\pi/2$. In this case, you get $\frac{2}{0}$. The fact that you get this tells you the answer is either going to be $+\infty$ if the left and right hand limits are both $+\infty$, $-\infty$ if the left and right hand limits are both $-\infty$, or it does not exist in the case that one of the left or right hand limits is $+\infty$ and the other is $-\infty$. You need to look at the sign of the numerator and denominator as $x \to \pi/2$ from the left and right separately to figure out what is going on. For example, as $x \to \frac{\pi}{2}^+$, i.e., from the right, $x$ is slightly larger than $\pi/2$. In that case, what is the sign of $\cos(2x)$ and $\cos(x)$, and therefore $\frac{1 - \cos(2x)}{2 \cos x}$???

For the second one, you "plug" in $0$ and get $\frac{0}{0}$. This is more complicated. This is called an indeterminate form because the answer could be anything depending on what functions are involved, and you need to do some sort of manipulation to figure out what the actual limit is, or if it exists.

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