lim _{x rightarrow 8} frac{sqrt{1+sqrt{1+x}}- | vedclass

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

Question

(C) To evaluate the limit $\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$,we rationalize the numerator:$\lim _{x \rightarrow 8} \frac{\sqr

Vedclass · Accepted Answer

$\frac{1}{24}$

Vedclass · Answer

(C) To evaluate the limit $\lim _{x ightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$,we rationalize the numerator:$\lim _{x ightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8} 	imes \frac{\sqrt{1+\sqrt{1+x}}+2}{\sqrt{1+\sqrt{1+x}}+2}$$= \lim _{x ightarrow 8} \frac{1+\sqrt{1+x}-4}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)}$$= \lim _{x ightarrow 8} \frac{\sqrt{1+x}-3}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)}$Now,rationalize the remaining radical in the numerator:$= \lim _{x ightarrow 8} \frac{\sqrt{1+x}-3}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)} 	imes \frac{\sqrt{1+x}+3}{\sqrt{1+x}+3}$$= \lim _{x ightarrow 8} \frac{1+x-9}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)(\sqrt{1+x}+3)}$$= \lim _{x ightarrow 8} \frac{x-8}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)(\sqrt{1+x}+3)}$$= \lim _{x ightarrow 8} \frac{1}{(\sqrt{1+\sqrt{1+x}}+2)(\sqrt{1+x}+3)}$Substitute $x = 8$:$= \frac{1}{(\sqrt{1+\sqrt{9}}+2)(\sqrt{9}+3)}$$= \frac{1}{(\sqrt{1+3}+2)(3+3)}$$= \frac{1}{(2+2)(6)} = \frac{1}{4 	imes 6} = \frac{1}{24}$

$\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$ is equal to

Similar Questions

$\mathop {\lim }\limits_{x \to \infty } \frac{{{{(x + 1)}^{10}} + {{(x + 2)}^{10}} + \dots + {{(x + 100)}^{10}}}}{{{x^{10}} + {{10}^{10}}}}$ is equal to

If $\lim _{x \rightarrow \infty}\left(\frac{11 x^3-3 x+4}{13 x^3-5 x^2-7}\right)=\frac{a}{b}$,then the value of $a+b$ equals:

The value of the limit $\lim _{x \rightarrow-\infty}\left(\sqrt{4 x^2-x}+2 x\right)$ is

The value of $\mathop {\lim }\limits_{x \to \infty } \frac{(x + 1)(3x + 4)}{x^2(x - 8)}$ is equal to

If $\alpha, \beta$ are the roots of the quadratic equation $ax^2 + bx + c = 0$,then $\lim_{x \to \alpha} \frac{1 - \cos(ax^2 + bx + c)}{(x - \alpha)^2}$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module