TS EAMCET · Maths · Limits
\(\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}\) is equal to
- A \(\frac{3}{2}\)
- B \(\frac{1}{4}\)
- C \(\frac{1}{24}\)
- D \(\frac{1}{12}\)
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}…
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