Could someone please explain the steps in the Maclauren series? 
Find the limit of $$\lim_{x\to 0}{{1-(\cos x)({\cos2x})^{1/2}(\cos3x)^{1/3}}\over{x^2}}$$

Steps to solving this problem, from the book:
Step 1: $$\lim_{x\to 0}{{1-(1-\frac{x^2}{2})({1-2x^2})^{1/2}(1-\frac{9x^2}{2})^{1/3}}\over{x^2}}$$
Step 2: $$\lim_{x\to 0}{{1-(1-\frac{x^2}{2})({1-x^2})(1-\frac{3x^2}{2})}\over{x^2}}$$
Step 3: $$\lim_{x\to 0}{{1-(1-\frac{3x^2}{2})(1-\frac{3x^2}{2})}\over{x^2}}$$
Step 4:  $$\lim_{x\to 0}{\frac{1-1+3x^2}{x^2}=3 }$$
Could someone please explain to me, what happened in step 2. How did we loose the roots ? 
And what happened from step 2 -> step 3 I have no idea what happened there ... :(
 A: It's the generalized binomial theorem.
In its simplest form,
it states that,
for fixed $a$,
$\lim_{x \to 0} \left(\dfrac{(1+x)^a-(1+ax)}{x} \right)
=0
$.
Informally,
$(1+x)^a \approx 1+ax
$
or,
more precisely,
in big-oh notation,
$(1+x)^a = 1+ax+O(x^2)
$.
This follows from
L'Hopital's rule because
$\lim_{x \to 0} \left(\dfrac{(1+x)^a-(1+ax)}{x} \right)
=\lim_{x \to 0} \left(\dfrac{a(1+x)^{a-1}-a}{1} \right)
=0
$.
A: You should probably get rid of the book. Most calculus texts are like this. The right approach is to use the little / big o notation.
The key here is to use Taylor / Maclaurin expansions as $x\to 0$ $$ \cos x=1-\frac{x^2}{2}+o(x^2)\tag{1}$$ and $$(1+x)^n=1+nx+o(x)\tag{2}$$ where $o(f(x)) $ represents a function $g(x) $ such that $g(x) /f(x) \to 0$.
The starting step (based on expansion $(1)$ above) thus needs to be written like $$\lim_{x\to 0}\dfrac{1-\left(1-\dfrac {x^2}{2}+o(x^2)\right)\left(1-2x^2+o(x^2)\right) ^{1/2}\left(1-\dfrac{9x^2}{2}+o(x^2)\right)^{1/3}} {x^2} $$ and the next step (using $(2)$) becomes $$\lim_{x\to 0}\dfrac{1-\left(1-\dfrac {x^2}{2}+o(x^2)\right)\left(1-x^2+o(x^2)\right)\left(1-\dfrac{3x^2}{2}+o(x^2)\right)} {x^2} $$ Finally we have via multiplication $$\lim_{x\to 0}\dfrac{1-\left(1-3x^2+o(x^2)\right)} {x^2}=3 $$

A simpler approach is to use standard limits $$\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2},\,\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$ which are equivalent to the Taylor expansions given earlier.
The numerator of the expression under limit is of the form $1-abc$ where $a, b, c$ all tend to $1$ as $x\to 0$. We can do a split like $$1-abc=1-a+a(1-bc)$$ and thus we have the desired limit as $$\lim_{x\to 0}\frac{1-\cos x} {x^2}+\lim_{x\to 0}\cos x\cdot\frac{1-(\cos 2x)^{1/2}(\cos 3x)^{1/3}} {x^2}$$ which is same as $$\frac{1}{2}+\lim_{x\to 0}\frac{1-(\cos 2x)^{1/2}(\cos 3x)^{1/3}}{x^2}$$ Applying the same technique we get $$\frac{1}{2}+\lim_{x\to 0}\frac{1-(\cos 2x)^{1/2}}{x^2}+\lim_{x\to 0}(\cos 2x)^{1/2}\cdot\frac{1-(\cos 3x)^{1/3}}{x^2}$$ The first limit above is evaluated as $$\lim_{x\to 0}\frac{1-(\cos 2x)^{1/2}}{1-\cos 2x}\cdot\frac{1-\cos 2x}{(2x)^2}\cdot 4=\frac{1}{2}\cdot\frac{1}{2}\cdot 4=1$$ and thus desired limit is equal to $$\frac{3}{2}+\lim_{x\to 0}\frac{1-(\cos 3x)^{1/3}}{x^2}$$ The last limit is evaluated in similar manner to get $3/2$ and thus the final answer is $3$.
This technique has been used for a more complicated problem here. 
