Question

$\displaystyle\lim_{x \to 0}\frac{1 - \cos 4x}{\tan^2 2x}$ is equal to

• A. 3
• B. $\dfrac{1}{4}$
• C. $\dfrac{1}{2}$
• D. 2
• E. 0

Hint:

A quick method: Use $1-\cos\theta \approx \dfrac{\theta^2}{2}$ and $\tan\theta \approx \theta$ near $\theta = 0$. So $\dfrac{1-\cos 4x}{\tan^2 2x} \approx \dfrac{(4x)^2/2}{(2x)^2} = \dfrac{8x^2}{4x^2} = 2$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Use standard limits: $\lim_{u\to 0}\dfrac{1-\cos u}{u^2} = \dfrac{1}{2}$ and $\lim_{u\to 0}\dfrac{\tan u}{u} = 1$.

Step 2: Detailed Explanation:
\[ \lim_{x\to 0}\frac{1-\cos 4x}{\tan^2 2x} = \lim_{x\to 0}\frac{1-\cos 4x}{(4x)^2}\cdot\frac{(2x)^2}{\tan^2 2x}\cdot\frac{(4x)^2}{(2x)^2} \] \[ = \frac{1}{2}\cdot 1\cdot \frac{16x^2}{4x^2} = \frac{1}{2}\cdot 4 = 2 \]

Step 3: Final Answer:
$\displaystyle\lim_{x\to 0}\frac{1-\cos 4x}{\tan^2 2x} = 2$.