Question

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

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

Answer 1

The correct option is B $\frac{1}{2}$

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

Let $y=\left(\sqrt{(1-\cos x)+\sqrt{(1-\cos x)+\sqrt{(1-\cos x)+\cdots \cdots \cdots \infty }}}\right)$

As $x\to 0$, we can replace $1-\cos x$ with $\frac{x^2}{2}$ as sinx can be approximated to x for small values of x
$\Rightarrow y^2-y=\frac{x^2}{2}$
$\Rightarrow 2y^2-2y-x^2=0$
As $y>0$,
$y=\frac{1+\sqrt{1+2x^2}}{2}$

$\therefore \underset{x\to 0}{lim}\frac{\left(\sqrt{(1-\cos x)+\sqrt{(1-\cos x)+\sqrt{(1-\cos x)+\cdots \cdots \cdots \infty }}}\right)-1}{x^2}$$=\underset{x\to 0}{lim}\frac{\sqrt{1+2x^2}-1}{2x^2}$
Rationalizing gives
$\underset{x\to 0}{lim}\frac{1}{\sqrt{1+2x^2}+1}$$=\frac{1}{2}$
Hence, option B is the correct answer.