:(lim) x 0 (1 - cos x * sqrt(cos 2x))/(tan^2 x)

1. Why use expansion/Taylor series?

When $x \to 0$, functions such as $\sin x$, $\cos x$, and $\tan x$ can be written as polynomials in $x$. These series give highly accurate values for tiny $x$, yet the algebra becomes easy.

Typical small–$x$ series (learn and remember):

What’s the problem?

We want to know what number the fraction

$\displaystyle\frac{1\; -\; \cos x\,\sqrt{\cos 2x}}{\tan^2 x}$

gets closer and closer to when $x$ himself gets very very tiny (moves toward $0$).

Picture it like this

• Think of $x$ as a super–small spice you are adding to a dish.
• The top of the fraction (numerator) is how the taste changes when you mix two ingredients, $\cos x$ and $\sqrt{\cos 2x}$, and then subtract the result from $1$.
• The bottom of the fraction (denominator) is $\tan^2 x$, also reacting to the same tiny spice.
• We want to see the ratio of these two changes.
  If both the top and bottom are shrinking to $0$, who shrinks faster? The answer to that gives the limit!

Big idea

Because $x$ is so small, we can replace fancy functions (cosine, tangent, square–root) by their baby–versions (first few terms of their Taylor series). That lets us compare them like simple polynomials and spot the final constant value quickly.

Step-by-step solution

1. Write the series for each function

   $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$

   $\cos 2x = 1 - 2x^2 + \frac{2}{3}x^4 + O(x^6)$

   Because we need $\sqrt{\cos 2x}$, set $y = \cos 2x - 1 = -2x^2 + \frac{2}{3}x^4$.

   $\sqrt{1+y} = 1 + \frac{y}{2} - \frac{y^2}{8} + O(y^3)$

   Substituting $y$ gives

   $\sqrt{\cos 2x} = 1 - x^2 - \frac{x^4}{6} + O(x^6)$

2. Multiply $\cos x$ and $\sqrt{\cos 2x}$

   \cos x \; \sqrt{\cos 2x} &=\left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right) \left(1 - x^2 - \frac{x^4}{6}\right)\\[6pt] &= 1 \; - \; \frac{3x^2}{2} \; + \; \frac{3x^4}{8} \; + O(x^6) \end{aligned}$$

3. Form the numerator

   $1 - \cos x\,\sqrt{\cos 2x} = \frac{3x^2}{2} - \frac{3x^4}{8} + O(x^6)$

4. Series for the denominator

   $\tan x = x + \frac{x^3}{3} + O(x^5)$

   $\tan^2 x = x^2 + \frac{2x^4}{3} + O(x^6)$

5. Take the ratio

   = \frac{\displaystyle \frac{3x^2}{2} - \frac{3x^4}{8} + O(x^6)}{\displaystyle x^2 + \frac{2x^4}{3} + O(x^6)}$$ Factor $x^2$ top and bottom: $$= \frac{\displaystyle \frac{3}{2} - \frac{3x^2}{8} + O(x^4)}{\displaystyle 1 + \frac{2x^2}{3} + O(x^4)}$$

6. Now let $x \to 0$

   The higher-power $x^2$ terms vanish, leaving

   $\boxed{\;\dfrac{3}{2}\;}$