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)

$\frac{1}{2}$

Substituting gives us indeterminate form.

Differentiating the numerator,

Let, $y=\sqrt{(1-\cos x+y)}$

$y^2=(1-\cos x+y)$

$2y{y}^{\prime }=\sin x+y^{\prime }$ so, $y^{\prime }=\frac{\sin x}{2y-1}$

Now Applying L'Hospitals Rule,

$={lim}_{x\to 0}\frac{\sin x}{(2y-1)2x}$

Substituting gives us indeterminate form, so applying L'Hospitals Rule,

$={lim}_{x\to 0}\frac{\cos x}{(4x{y}^{\prime }+4y-2)}$

$=\frac{1}{2}(y=1$ when $x=0)$

Hence, option 'B' is correct.