Question

$\lim_{x\to0} \frac{x\cot(4x)}{\sin^2x \cot^2(2x)}$ is equal to:

• A. 1
• B. 4
• C. 0
• D. 2

Answer 1

Answer: (A)

1

Answer 2

To evaluate the limit $\lim_{x\to0} \frac{x\cot(4x)}{\sin^2x \cot^2(2x)}$, we first rewrite the cotangent terms in terms of tangent, as $\cot \theta = \frac{1}{\tan \theta}$.

The expression becomes: $\lim_{x\to0} \frac{x \cdot \frac{1}{\tan(4x)}}{\sin^2x \cdot \frac{1}{\tan^2(2x)}}$ Simplify the expression: $\lim_{x\to0} \frac{x \tan^2(2x)}{\sin^2x \tan(4x)}$ Now, we use the standard limits: $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \quad \text{and} \quad \lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$ To apply these limits, we multiply and divide each trigonometric term by its argument.

Let's rewrite the numerator and the denominator:

Numerator: $x \tan^2(2x)$ $x \tan^2(2x) = x \cdot \left(\frac{\tan(2x)}{2x} \cdot 2x\right)^2$ $= x \cdot \left(\frac{\tan(2x)}{2x}\right)^2 \cdot (2x)^2$ $= x \cdot \left(\frac{\tan(2x)}{2x}\right)^2 \cdot 4x^2$ $= 4x^3 \left(\frac{\tan(2x)}{2x}\right)^2$

Denominator: $\sin^2x \tan(4x)$ $\sin^2x \tan(4x) = \left(\frac{\sin x}{x} \cdot x\right)^2 \cdot \left(\frac{\tan(4x)}{4x} \cdot 4x\right)$ $= \left(\frac{\sin x}{x}\right)^2 \cdot x^2 \cdot \left(\frac{\tan(4x)}{4x}\right) \cdot 4x$ $= 4x^3 \left(\frac{\sin x}{x}\right)^2 \left(\frac{\tan(4x)}{4x}\right)$

Now substitute these modified terms back into the limit expression: $\lim_{x\to0} \frac{4x^3 \left(\frac{\tan(2x)}{2x}\right)^2}{4x^3 \left(\frac{\sin x}{x}\right)^2 \left(\frac{\tan(4x)}{4x}\right)}$ Cancel out the common term $4x^3$: $\lim_{x\to0} \frac{\left(\frac{\tan(2x)}{2x}\right)^2}{\left(\frac{\sin x}{x}\right)^2 \left(\frac{\tan(4x)}{4x}\right)}$ Now, apply the limits as $x \to 0$: $\lim_{x\to0} \frac{\tan(2x)}{2x} = 1$ $\lim_{x\to0} \frac{\sin x}{x} = 1$ $\lim_{x\to0} \frac{\tan(4x)}{4x} = 1$ Substitute these values into the expression: $\frac{(1)^2}{(1)^2 \cdot (1)} = \frac{1}{1 \cdot 1} = 1$

Thus, the limit is 1.