$\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$ is equal to
$\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$ is equal to
- $\frac{3}{2}$
- $\frac{1}{4}$
- $\frac{1}{24}$
- $\frac{1}{12}$
Solution
$\begin{aligned} & \lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8} \times \frac{\sqrt{1+\sqrt{1+x}}+2}{\sqrt{1+\sqrt{1+x}}+2} \\ & =\lim _{x \rightarrow 8} \frac{1+\sqrt{1+x}-4}{(\sqrt{1+\sqrt{1+x}}+2)(x-8)} \\ & =\lim _{x \rightarrow 8} \frac{\sqrt{1+x}-3}{(\sqrt{1+\sqrt{1+x}}+2)(x-8)} \times \frac{\sqrt{1+x}+3}{(\sqrt{1+x}+3)} \\ & =\lim _{x \rightarrow 8} \frac{1+x-9}{(\sqrt{1+x}+3)(\sqrt{1+\sqrt{1+x}}+2)(x-8)} \\ & =\lim _{x \rightarrow 8} \frac{1}{(\sqrt{1+x}+3)(\sqrt{1+\sqrt{1+x}}+2)} \\ & =\frac{1}{(\sqrt{1+8}+3)(\sqrt{1+\sqrt{1+8}}+2)} \\ & =\frac{1}{(3+3)(2+2)}=\frac{1}{24}\end{aligned}$
Asked in: AP EAMCET 2011
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