Solve $\displaystyle\lim_{x \to 0}{(1 - \sqrt{\cos 4x})\cot^2x}$

Question

I just use L'Hospital's Rule (by changing into $\dfrac{1}{\tan^2x}$ first) for this, but it makes me even more confusing to solve. Could anyone help me to figure this out please?

Answer 1

Note,

$$1-\sqrt{\cos4x}={1-\cos4x\over1+\sqrt{\cos4x}}={1-(\cos^22x-\sin^22x)\over1+\sqrt{\cos4x}}={2\sin^22x\over1+\sqrt{\cos4x}}={8\sin^2x\cos^2x\over1+\sqrt{\cos4x}}$$

Consequently, for the limit as $x\to0$, we have

$$(1-\sqrt{\cos4x})\cot^2x={8\sin^2x\cos^2x\over1+\sqrt{\cos4x}}\cdot{\cos^2x\over\sin^2x}={8\cos^4x\over1+\sqrt{\cos4x}}\to{8\cdot1^4\over1+\sqrt1}={8\over2}=4$$

The key thing here, really, is that multiplying by the conjugate $1+\sqrt{\cos4x}$ eliminates any difficulty with the problematic square root, since $1+\sqrt{\cos4x}$ tends to a nice non-zero limit.

Answer 2

HINT

$$\lim_{x\to 0} \left(1-\sqrt{\cos(4x)}\right)\cot^2{x}\\=\lim_{x\to 0}\frac{1-\cos^2{4x}}{\tan^2{x}\cdot \left(1+\sqrt{\cos(4x)}\right)\cdot \left(1+\cos(4x)\right)}\\=\lim_{x\to 0}\frac{\sin^2{4x}}{\tan^2{x}}\cdot \lim_{x\to 0}\frac{1}{\left(1+\sqrt{\cos(4x)}\right)}\cdot\lim_{x\to 0}\frac{1} {\left(1+\cos(4x)\right)}$$

Answer 3

HINT

By standard limits

$$(1 - \sqrt{\cos 4x})(\cot^2x)=\frac{1 - \sqrt{\cos 4x}}{\tan^2x}=\frac{1 - \cos 4x}{(4x)^2}\cdot \frac{x^2}{\tan^2x}\cdot\frac{16}{1 + \sqrt{\cos 4x}}$$