mathop {lim }limits_{x to infty } frac{{sqrt | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $f(x) = \sqrt{x^{2} - \sqrt{x^{2} - \sqrt{x^{2} - \dots}}}$Then $f(x) = \sqrt{x^{2} - f(x)}$.Squaring both sides,we get $f(x)^{2} = x^{2} - f(

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Let $f(x) = \sqrt{x^{2} - \sqrt{x^{2} - \sqrt{x^{2} - \dots}}}$Then $f(x) = \sqrt{x^{2} - f(x)}$.Squaring both sides,we get $f(x)^{2} = x^{2} - f(x)$,which implies $f(x)^{2} + f(x) - x^{2} = 0$.Using the quadratic formula for $f(x)$,we have $f(x) = \frac{-1 \pm \sqrt{1 + 4x^{2}}}{2}$.Since $f(x) > 0$,we take the positive root: $f(x) = \frac{-1 + \sqrt{1 + 4x^{2}}}{2}$.Now,we evaluate the limit: $\mathop {\lim }\limits_{x 	o \infty } \frac{f(x)}{x} = \mathop {\lim }\limits_{x 	o \infty } \frac{-1 + \sqrt{1 + 4x^{2}}}{2x}$.Dividing numerator and denominator by $x$,we get $\mathop {\lim }\limits_{x 	o \infty } \frac{-\frac{1}{x} + \sqrt{\frac{1}{x^{2}} + 4}}{2} = \frac{0 + \sqrt{0 + 4}}{2} = \frac{2}{2} = 1$.

$\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {{x^2} - \sqrt {{x^2} - \sqrt {{x^2} - \dots} } } }}{x}$ is equal to-

Similar Questions

$\lim _{n \rightarrow \infty}\left[\frac{1^3}{1-n^4}+\frac{2^3}{1-n^4}+\ldots +\frac{n^3}{1-n^4}\right]=$

$\mathop {\lim }\limits_{x \to \frac{\pi^+}{2}} e^{[\cot x]}$ is equal to :-
(where $[.]$ is the greatest integer function)

$\mathop {\lim }\limits_{n \to \infty } \frac{{n{{(2n + 1)}^2}}}{{(n + 2)({n^2} + 3n - 1)}} = $

$\mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^{ - 1/3}}}}{{1 - {x^{ - 2/3}}}} = $

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} (\text{cosec}\,x - \cot x)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module