Evaluation of limits using McLaurin series

Question

I have some questions regarding use of McLaurin series for evaluating limits. I stumbled upon a problem and I'm stuck.

Here is the problem:

$$ \lim_{x\to 0} \frac{1-\cos(x)(\cos(2x))^{1/2}}{x^{2}}$$

I expand first cosine to the second power and I expand second one to the fourth power of x. I'm stuck on the next step since i cannot factor out x squared from both expressions because I get $1/x^{2}$ and when x goes to zero that apparently isn't what I should get in the answer. Am I not seeing something obvious here?

Answer 1

We need the following expansion for $x\to0$

$$\cos x=1-\frac{x^2}{2}+o(x^2)$$

$$\cos 2x=1-\frac{4x^2}{2}+o(x^2)=1-2x^2+o(x^2)$$

$$(1+x)^a=1+ax+o(x)\implies \sqrt{\cos 2x}=(1-2x^2+o(x^2))^\frac12=1-x^2+o(x^2)$$

thus

$$\frac{1-\cos{x}\sqrt{\cos2x}}{x^2}=\frac{1-\left(1-\frac{x^2}{2}+o(x^2)\right)(1-x^2+o(x^2))}{x^2}=$$ $$\frac{1-1+\frac{3x^2}{2}+o(x^2)}{x^2}=\frac32+o(1)\to \frac32$$

Answer 2

$$\cos{x}=1-\frac{x^2}{2}+...,$$ $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+....$$ Thus, $$\frac{1-\cos{x}\sqrt{\cos2x}}{x^2}=\frac{1-\left(1-\frac{x^2}{2}+...\right)\left(1+\frac{-2x^2+...}{2}\right)}{x^2}\rightarrow\frac{3}{2}$$

Answer 3

$$\cos(x) \approx 1 - \frac{x^2}{2}$$

$$\cos(2x) \approx 1 - 2x^2$$

Substitute and you get

$$\frac{1 - (1 - x^2/2)\sqrt{1 - 2x^2}}{x^2}$$

Another expansion for the square root:

$$\sqrt{1 - 2x^2}\approx 1-x^2$$

Hence

$$\frac{1 - (1 - x^2/2)(1 - x^2)}{x^2} = \frac{1 - (1 - x^2 - x^2/2 + x^4/2)}{x^2} = \frac{3x^2/2 - x^4/2}{x^2} = \frac{3}{2}$$

Answer 4

Hint: $$\lim_{x\to 0} \frac{1-\cos(x)(\cos(2x))^{1/2}}{x^{2}}=\lim_{x\to 0} \frac{1-\cos(x)(\cos(2x))^{1/2}}{x^{2}}\frac{1+\cos(x)(\cos(2x))^{1/2}}{1+\cos(x)(\cos(2x))^{1/2}}\\=\lim_{x \to 0}\frac{1-\cos^2(x)\cos(2x)}{x^{2}}\frac{1}{1+\cos(x)(\cos(2x))^{1/2}}$$

The second limit is a continuous function, and the first can be evaluated with MacLaurin series.