MCQ
$\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {{x^2} - \sqrt {{x^2} - \sqrt {{x^2} - .....} } } }}{x}$ =
- A$0$
- B$\frac{1}{2}$
- C$1$
- D$\frac{1}{4}$
$\Rightarrow f^{2}(\mathrm{x})+f(\mathrm{x})=\mathrm{x}^{2}$
$\Rightarrow\left(f(x)+\frac{1}{2}\right)^{2}=x^{2}+\frac{1}{4}$
clearly
$x < f\left( x \right) + \frac{1}{2} < x + \frac{1}{2}$
$ \Rightarrow x - \frac{1}{2} < f\left( x \right) < x$
$\Rightarrow 1-\frac{1}{2 x}<\frac{f(x)}{x}<1$
$\Rightarrow$ by sandwith theorem $\mathop {\lim }\limits_{x \to \infty } \frac{{f(x)}}{x} = 1.$
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