Question

$\underset{x\to 0}{lim}\frac{\left(\sqrt{(1-\cos x)+\sqrt{(1-\cos x)+\sqrt{(1-\cos x)+......\infty }}}\right)-1}{x^2}$ equals.

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

Answer: (B)

$\frac{1}{2}$

Let $y=\sqrt{(1-\cos x)+\sqrt{(1-\cos x)+.....\infty }}$
$y=\sqrt{(1-\cos x)+y}$
$y^2=(1-\cos x)+y$
$y^2-y+(\cos x-1)=0$
$y=\frac{1\pm \sqrt{1-4\cos x+4}}{2}=\frac{1\pm \sqrt{5-4\cos x}}{2}$
$L=\underset{x\to 0}{lim}\frac{\frac{1\pm \sqrt{5-4\cos x}}{2}-1}{x^2}=\underset{x\to 0}{lim}\frac{\pm \sqrt{5-4\cos x}-1}{2x^2}$
Using L Hospital's Rule
$L=\underset{x\to 0}{lim}\pm \frac{1\times 4\sin x}{2\sqrt{5-4\cos x}\times 4x}$
$={lim}_{x\to 0}\pm \frac{1}{2\sqrt{5-4\cos x}}\times \frac{\sin x}{x}$
$=\frac{1}{2}$