lim x → 0 (1-cos x.(cos 2x)^1/2)/tan ^2x

1. Spot the indeterminate form

As $x \to 0$:

• $\cos x \to 1$ and $\cos 2x \to 1$ (\Rightarrow) the numerator tends to $1-1=0$.
• $\tan x \to 0$ (\Rightarrow) the denominator tends to $0$. So the whole fraction is of the form $\tfrac{0}{0}$, which is indeterminate and a green light for series expansion (or L’Hospital, but series is usually quicker in JEE).

2. Use Maclaurin/Taylor series for small $x$

For small angles (in radians!):

Because we see a square root of $\cos 2x$, we also need the expansion of $\sqrt{1+s}$. For small $s$,

$\sqrt{1+s}\approx 1+\frac{s}{2}-\frac{s^2}{8}+\dots$

3. Logical chain of thought

1. Write $\cos 2x=1+s$ with $s=-2x^2+\tfrac{2}{3}x^4+\dots$
2. Plug $s$ into the square-root formula to get $\sqrt{\cos 2x}$ up to $x^4$.
3. Multiply $\cos x$ and $\sqrt{\cos 2x}$ – keep terms up to $x^4$ only (because denominator will have $x^2$ and higher, so any term beyond $x^4$ will vanish in the limit).
4. Form the numerator $1-\bigl( \cos x\,\sqrt{\cos 2x}\bigr)$. Notice the first surviving term is proportional to $x^2$.
5. Expand $\tan^2 x$ up to $x^4$.
6. Factor $x^2$ from both top and bottom, cancel, and safely put $x=0$ – the leftover constant is the required limit.

Each of these steps is chosen because we only need as many terms as required to cancel the smallest powers of $x$ that give the first non-zero constant.

🤔 What’s the problem about?

We want to know what happens to a fraction when $x$ gets really, really close to $0$.
The fraction looks scary:

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

🪄 How do we tame it?

1. Think "tiny $x$" – when $x$ is tiny, sine, cosine, tangent, etc. can be replaced by very simple approximate numbers (their series).
2. Top and bottom both become 0 when $x\to0$ – that 0/0 form is a hint to expand them and cancel the common small pieces.
3. After cancelling, we just look at the first non-zero numbers left.

That’s really it – expand, cancel, read the leftover number. 🎉

Step-by-step Solution

1. Series needed (up to $x^4$):

   $$\begin{aligned} \cos x & = 1-\frac{x^2}{2}+\frac{x^4}{24}+\dots \\[4pt] \cos 2x & = 1-2x^2+\frac{2}{3}x^4+\dots \\[4pt] \tan x & = x+\frac{x^3}{3}+\dots \;\Longrightarrow\; \tan^2 x = x^2+\frac{2}{3}x^4+\dots \end{aligned}$$

2. Square root of $\cos 2x$: write $\cos 2x = 1+s$ with $s=-2x^2+\tfrac{2}{3}x^4$. Then

   $$\sqrt{\cos 2x}=1+\frac{s}{2}-\frac{s^2}{8}+\dots$$

   Compute: [ s = -2x^2 + \frac{2}{3}x^4 \quad\Rightarrow\quad \frac{s}{2} = -x^2 + \frac{1}{3}x^4 ] [ s^2 = \bigl(-2x^2\bigr)^2 + \text{higher} = 4x^4 \quad\Rightarrow\quad -\frac{s^2}{8} = -\frac{4x^4}{8} = -\frac{x^4}{2} ] Therefore

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

3. Product $\cos x \times \sqrt{\cos 2x}$ (keep terms up to $x^4$):

   $$\begin{aligned} &\bigl(1-\tfrac{x^2}{2}+\tfrac{x^4}{24}\bigr)\bigl(1-x^2-\tfrac{1}{6}x^4\bigr) \\[4pt] =\;& 1 -\bigl(x^2+\tfrac{x^2}{2}\bigr) + \bigl(-\tfrac{1}{6}+\tfrac{1}{2}+\tfrac{1}{24}\bigr)x^4 + \dots \\[4pt] =\;& 1 - \tfrac{3}{2}x^2 + \tfrac{3}{8}x^4 + \dots \end{aligned}$$

4. Numerator:

   $$1 - \cos x\,\sqrt{\cos 2x} = 1-\Bigl[1 - \tfrac{3}{2}x^2 + \tfrac{3}{8}x^4\Bigr] = \tfrac{3}{2}x^2 - \tfrac{3}{8}x^4 + \dots$$

5. Denominator:

   $$\tan^2 x = x^2 + \tfrac{2}{3}x^4 + \dots$$

6. Form the fraction, cancel $x^2$:

   $$\frac{\tfrac{3}{2}x^2 - \tfrac{3}{8}x^4}{x^2 + \tfrac{2}{3}x^4} = \frac{\tfrac{3}{2} - \tfrac{3}{8}x^2}{1 + \tfrac{2}{3}x^2}$$

7. Take the limit $x \to 0$ (all $x^2$ terms vanish):

   $$\boxed{\displaystyle \frac{3}{2}}$$