lim x → 0 (1-cos x.cos 2x)/tan ^2x

Key Ideas to Crack Such Limits

1. Standard small-angle expansions (also called Maclaurin/Taylor series for trigonometric functions):
   • $\sin x \approx x - \frac{x^3}{6}$
   • $\cos x \approx 1 - \frac{x^2}{2}$
   • $\tan x \approx x + \frac{x^3}{3}$

2. Order of smallness: For $x \to 0$, the lowest power of $x$ dominates. If both numerator and denominator behave like $x^2$, the ratio tends to a finite number.

3. Product Rule for expansions: When you have $\cos x \cdot \cos 2x$, expand each and multiply, keeping only the most significant (lowest-power) terms.

4. Why not L'Hôpital? You could differentiate twice, but series is often faster and cleaner for JEE-type problems.

Logical Chain a Student Should Follow

1. Recognise an indeterminate form $\frac{0}{0}$.
2. Decide between L'Hôpital and series. For trigonometric limits at 0, series is generally quicker.
3. Expand $\cos x$ and $\cos 2x$ individually to $x^2$ order.
4. Multiply the expansions to find the numerator’s smallest non-zero term.
5. Expand $\tan x$ to first order because squaring will raise it to $x^2$.
6. Cancel the common $x^2$ factor and simplify.
7. Write the final constant value.

Imagine very small angles

When $x$ is super-tiny, angles act a bit like straight lines:

• $\cos(\text{tiny angle})$ is almost 1, but slips down a little.
• $\tan(\text{tiny angle})$ is almost the angle itself.

The question asks:

As x goes to 0, what value does
(1 – cos x · cos 2x) / (tan x)^2
settle on?

Think of "1 – cos x · cos 2x" as how much the two cosines together fall below 1. For tiny $x$ that fall is proportional to $x^2$. Meanwhile $(\tan x)^2$ is also roughly $x^2$. So both the top and bottom shrink at the same speed. Divide their sizes and the $x^2$ bits cancel, leaving a neat constant: 2.5 (which is $\frac{5}{2}$).

Step-by-Step Solution

1. Write the limit $L = \lim_{x \to 0} \frac{1- \cos x \; \cos 2x}{\tan^2 x}$

2. Small-angle expansions (up to $x^2$) [ \cos x \approx 1 - \frac{x^2}{2}, \qquad \cos 2x \approx 1 - 2x^2 ]

3. Multiply them (ignore terms $\ge x^4$) [ \cos x ; \cos 2x \approx \left(1 - \frac{x^2}{2}\right)\left(1 - 2x^2\right) ] [ \approx 1 - \frac{x^2}{2} - 2x^2 \quad (\text{drop } x^4 \text{ term}) ] [ = 1 - \frac{5x^2}{2} ]

4. Numerator [ 1 - \cos x, \cos 2x \approx 1 - \left(1 - \frac{5x^2}{2}\right) = \frac{5x^2}{2} ]

5. Denominator $\tan x \approx x \;\Rightarrow\; \tan^2 x \approx x^2$

6. Form the ratio $\frac{\tfrac{5x^2}{2}}{x^2} = \frac{5}{2}$

7. Take the limit $L = \boxed{\dfrac{5}{2}}$