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

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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{\sq

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 numerator again: $= \lim _{x ightarrow 8} \frac{\sqrt{1+x}-3}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)} 	imes \frac{\sqrt{1+x}+3}{\sqrt{1+x}+3}$ $= \lim _{x ightarrow 8} \frac{1+x-9}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)(\sqrt{1+x}+3)}$ $= \lim _{x ightarrow 8} \frac{x-8}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)(\sqrt{1+x}+3)}$ $= \lim _{x ightarrow 8} \frac{1}{(\sqrt{1+\sqrt{1+x}}+2)(\sqrt{1+x}+3)}$ Substitute $x = 8$: $= \frac{1}{(\sqrt{1+\sqrt{9}}+2)(\sqrt{9}+3)} = \frac{1}{(\sqrt{4}+2)(3+3)} = \frac{1}{(2+2)(6)} = \frac{1}{4 	imes 6} = \frac{1}{24}$

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

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