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STD 11 - 12. limits — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 12. limits4 Marks
MCQ
$\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {{x^2} - \sqrt {{x^2} - \sqrt {{x^2} - .....} } } }}{x}$ is equal to-
  • A
    $0$
  • B
    $\frac{1}{2}$
  • ✓
    $1$
  • D
    $\frac{1}{4}$
✓

Answer

Correct option: C.
$1$
c
Let $f(x)=\sqrt{x^{2}-\sqrt{x^{2}-\sqrt{x^{2}} \ldots .}}$

$\Rightarrow f^{2}(\mathrm{x})+f(\mathrm{x})=\mathrm{x}^{2}$

$\Rightarrow\left(f(x)+\frac{1}{2}\right)^{2}=x^{2}+\frac{1}{4}$

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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