Question

If $L=\underset{\theta \to 0}{lim}\frac{co{s}^2(1-co{s}^2(1-co{s}^2(......(1-co{s}^2\theta \left)\right)\left)\right)}{sin\left(\frac{\pi (\sqrt{\theta +4}-2)}{\theta }\right)}$exists and takes a non zero value, then find $L$.

• A. $\sqrt{2}$
• B. $-\sqrt{2}$
• C. $1$
• D. $-1$

Answer 1

Answer: (A)

$\sqrt{2}$

$\underset{\theta \to 0}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{\theta \to 0}{lim}\frac{co{s}^2(1-co{s}^2(1-co{s}^2(1-co{s}^2\theta )\left)\right)}{sin\left(\frac{\pi (\sqrt{\theta +4}-2)}{\theta }\right)}$
${lim}_{\theta \to 0}f\left(x\right)=1$, and ${lim}_{\theta \to 0}g\left(x\right)=sin\left(\frac{0}{0}\right)$
Using the quotient rule of limits:
If ${lim}_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}$ exists and ${lim}_{x\to a}f\left(x\right)$ also exists, then ${lim}_{x\to a}g\left(x\right)$ have to exist.
$\therefore {lim}_{\theta \to 0}sin\left(\frac{\pi (\sqrt{\theta +4}-2)}{\theta }\right)=sin\left({lim}_{\theta \to 0}\frac{\pi (\sqrt{\theta +4}-2)}{\theta }\right)$
$\Rightarrow sin\left({lim}_{\theta \to 0}\frac{\pi }{\sqrt{4+\theta }+2}\right)=sin\left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$
$\therefore L=\sqrt{2}$