ExamBro
ExamBro
TS EAMCET · Maths · Limits

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

  1. A \(\frac{3}{2}\)
  2. B \(\frac{1}{4}\)
  3. C \(\frac{1}{24}\)
  4. D \(\frac{1}{12}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{24}\)

Step-by-step Solution

Detailed explanation

\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}…